What does typedef with no type mean?












3















From the Standard N1570 6.7.8:




A typedef declaration does not introduce a new type, only a synonym
for the type so specified.




So I expected that it is not possible to write something like this:



typedef t;

t *t_ptr;


and it should fail to compile since no type to introduce a synonym to provided. But it is fine: Demo. So what does this ever mean and why does it compile?






c typedef







share|improve this question




















  • 3





    It is as fine as MSVC is a fine C compiler, i.e. not fine at all.

    – Antti Haapala
    Nov 28 '18 at 5:33
















3















From the Standard N1570 6.7.8:




A typedef declaration does not introduce a new type, only a synonym
for the type so specified.




So I expected that it is not possible to write something like this:



typedef t;

t *t_ptr;


and it should fail to compile since no type to introduce a synonym to provided. But it is fine: Demo. So what does this ever mean and why does it compile?






c typedef







share|improve this question




















  • 3





    It is as fine as MSVC is a fine C compiler, i.e. not fine at all.

    – Antti Haapala
    Nov 28 '18 at 5:33














3












3








3


1






From the Standard N1570 6.7.8:




A typedef declaration does not introduce a new type, only a synonym
for the type so specified.




So I expected that it is not possible to write something like this:



typedef t;

t *t_ptr;


and it should fail to compile since no type to introduce a synonym to provided. But it is fine: Demo. So what does this ever mean and why does it compile?






c typedef







share|improve this question
















From the Standard N1570 6.7.8:




A typedef declaration does not introduce a new type, only a synonym
for the type so specified.




So I expected that it is not possible to write something like this:



typedef t;

t *t_ptr;


and it should fail to compile since no type to introduce a synonym to provided. But it is fine: Demo. So what does this ever mean and why does it compile?






c typedef




c typedef






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited Nov 29 '18 at 20:40









dbush

102k13108144




102k13108144










asked Nov 28 '18 at 5:16









Some NameSome Name

1,517417




1,517417








  • 3





    It is as fine as MSVC is a fine C compiler, i.e. not fine at all.

    – Antti Haapala
    Nov 28 '18 at 5:33














  • 3





    It is as fine as MSVC is a fine C compiler, i.e. not fine at all.

    – Antti Haapala
    Nov 28 '18 at 5:33








3




3





It is as fine as MSVC is a fine C compiler, i.e. not fine at all.

– Antti Haapala
Nov 28 '18 at 5:33





It is as fine as MSVC is a fine C compiler, i.e. not fine at all.

– Antti Haapala
Nov 28 '18 at 5:33












3 Answers
3






active

oldest

votes


















6














This relies on the fact that, missing type specification defaults to int.



So, your statement



 typedef t;


is the same as



 typedef int t;


With the proper level of warning, compiler emits warning:



warning: type defaults to ‘int’ in declaration of ‘t’ [-Wimplicit-int]

 typedef t;

         ^


That said, do not rely on this behaviour, "implicit int" rule has been obsolete since C99.






share|improve this answer


























  • So reasonable to ask if it is Standardized?

    – Some Name
    Nov 28 '18 at 5:23











  • @SomeName it was, 3 decades ago and has not been for 2 decades.

    – Antti Haapala
    Nov 28 '18 at 5:25











  • @AnttiHaapala So now the behavior is undefined or implementation defined?

    – Some Name
    Nov 28 '18 at 5:27






  • 2





    @SomeName neither. A C99, C11, C17 compiler must diagnose this as an incorrect program.

    – Antti Haapala
    Nov 28 '18 at 5:28













  • @SomeName The support itself for implicit int is removed, though it is still provided as compiler extension. Use strict mode , and your code should fail to compile in first place.

    – Sourav Ghosh
    Nov 28 '18 at 5:29





















3














It defaults to an int.



The compiler warning shows what is going on: 



#1 with x86-64 gcc 8.2

<source>:1:9: warning: type defaults to 'int' in declaration of 't' [-Wimplicit-int]

 typedef t;


From C99 onwards, the implicit int rule was removed. So this is not applicable from C99 onward.



If you use the -pedantic-errors compiler option in GCC (meaning strict conformity to the standard), it issues an error. See here.



If you are interested, the relevant section in C89 standard which allowed this:




3.5.2 Type specifiers

Each list of type specifiers shall be one of the following sets; the type specifiers may occur in any order, possibly intermixed with the other declaration specifiers.




  • void

  • char

  • signed char

  • unsigned char

  • short , signed short , short int , or signed short int

  • unsigned short , or unsigned short int

  • int , signed , signed int , or no type specifiers




So in C99, the last part of what was bolded above (or no type specifiers) was removed.






share|improve this answer





















  • 1





    Please don't confuse readers by quoting now-outdated C89, not everyone may know about the applicability of the standards. :)

    – Sourav Ghosh
    Nov 28 '18 at 5:39













  • I did mention that it is not relevant from C99 onwards. But I can make it extra clear.

    – P.W
    Nov 28 '18 at 5:40











  • I would really appreciate if you move the positioning of the two blocks. First, mention about C99/C11 and then, about C89. Just my two cents.

    – Sourav Ghosh
    Nov 28 '18 at 5:45











  • @SouravGhosh: Done.

    – P.W
    Nov 28 '18 at 5:48



















0














A typedef declaration defines a synonym for an object or pointer type. So you should specify both the type for which you want to create a synonym, and a name to use as synonym.



For example:



// 'byte_t' is a synonym for 'unsigned char'

typedef unsigned char byte_t;

// 'handler_t' is a synonym for 'void (*)(int)', a function pointer

typedef void (*handler_t)(int);

// 'short_p' is a synonym for 'short *'

typedef short * short_p;

// 'record' is a synonym for an anonymous structure

typedef struct {

    int a;

    char *b;

    } record;

// 'error_p' is a synonym for a pointer to 'struct error', defined somewhere else

typedef struct error *error_p;


Many more examples in the source you cited.






share|improve this answer





















  • 1





    This is fine, but how does this answers the question?

    – Sourav Ghosh
    Nov 28 '18 at 5:55











  • @Sourav Ghosh I felt that the typedef syntax was new to him, so I give him some basic examples with comments. Other answers already addressed the compilation problem in itself.

    – A. Semat
    Nov 28 '18 at 6:07











  • Please don't invent your own terms and grammar that don't even exist in the C language. There are no syntax items called "base-type" or "type-alias".

    – Lundin
    Nov 28 '18 at 9:53













  • @Lundin. Thanks. I should have specified that base-type and type-alias should have been replaced by actual types and names. Moreover that pseudo-syntax didn't suit function types. I also changed alias to synonym, since the C standards refers to typedef names as such.

    – A. Semat
    Nov 28 '18 at 10:29











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3 Answers
3






active

oldest

votes








3 Answers
3






active

oldest

votes









active

oldest

votes






active

oldest

votes









6














This relies on the fact that, missing type specification defaults to int.



So, your statement



 typedef t;


is the same as



 typedef int t;


With the proper level of warning, compiler emits warning:



warning: type defaults to ‘int’ in declaration of ‘t’ [-Wimplicit-int]

 typedef t;

         ^


That said, do not rely on this behaviour, "implicit int" rule has been obsolete since C99.






share|improve this answer


























  • So reasonable to ask if it is Standardized?

    – Some Name
    Nov 28 '18 at 5:23











  • @SomeName it was, 3 decades ago and has not been for 2 decades.

    – Antti Haapala
    Nov 28 '18 at 5:25











  • @AnttiHaapala So now the behavior is undefined or implementation defined?

    – Some Name
    Nov 28 '18 at 5:27






  • 2





    @SomeName neither. A C99, C11, C17 compiler must diagnose this as an incorrect program.

    – Antti Haapala
    Nov 28 '18 at 5:28













  • @SomeName The support itself for implicit int is removed, though it is still provided as compiler extension. Use strict mode , and your code should fail to compile in first place.

    – Sourav Ghosh
    Nov 28 '18 at 5:29


















6














This relies on the fact that, missing type specification defaults to int.



So, your statement



 typedef t;


is the same as



 typedef int t;


With the proper level of warning, compiler emits warning:



warning: type defaults to ‘int’ in declaration of ‘t’ [-Wimplicit-int]

 typedef t;

         ^


That said, do not rely on this behaviour, "implicit int" rule has been obsolete since C99.






share|improve this answer


























  • So reasonable to ask if it is Standardized?

    – Some Name
    Nov 28 '18 at 5:23











  • @SomeName it was, 3 decades ago and has not been for 2 decades.

    – Antti Haapala
    Nov 28 '18 at 5:25











  • @AnttiHaapala So now the behavior is undefined or implementation defined?

    – Some Name
    Nov 28 '18 at 5:27






  • 2





    @SomeName neither. A C99, C11, C17 compiler must diagnose this as an incorrect program.

    – Antti Haapala
    Nov 28 '18 at 5:28













  • @SomeName The support itself for implicit int is removed, though it is still provided as compiler extension. Use strict mode , and your code should fail to compile in first place.

    – Sourav Ghosh
    Nov 28 '18 at 5:29
















6












6








6







This relies on the fact that, missing type specification defaults to int.



So, your statement



 typedef t;


is the same as



 typedef int t;


With the proper level of warning, compiler emits warning:



warning: type defaults to ‘int’ in declaration of ‘t’ [-Wimplicit-int]

 typedef t;

         ^


That said, do not rely on this behaviour, "implicit int" rule has been obsolete since C99.






share|improve this answer















This relies on the fact that, missing type specification defaults to int.



So, your statement



 typedef t;


is the same as



 typedef int t;


With the proper level of warning, compiler emits warning:



warning: type defaults to ‘int’ in declaration of ‘t’ [-Wimplicit-int]

 typedef t;

         ^


That said, do not rely on this behaviour, "implicit int" rule has been obsolete since C99.







share|improve this answer














share|improve this answer



share|improve this answer








edited Nov 28 '18 at 5:52

























answered Nov 28 '18 at 5:22









Sourav GhoshSourav Ghosh

111k15131190




111k15131190













  • So reasonable to ask if it is Standardized?

    – Some Name
    Nov 28 '18 at 5:23











  • @SomeName it was, 3 decades ago and has not been for 2 decades.

    – Antti Haapala
    Nov 28 '18 at 5:25











  • @AnttiHaapala So now the behavior is undefined or implementation defined?

    – Some Name
    Nov 28 '18 at 5:27






  • 2





    @SomeName neither. A C99, C11, C17 compiler must diagnose this as an incorrect program.

    – Antti Haapala
    Nov 28 '18 at 5:28













  • @SomeName The support itself for implicit int is removed, though it is still provided as compiler extension. Use strict mode , and your code should fail to compile in first place.

    – Sourav Ghosh
    Nov 28 '18 at 5:29





















  • So reasonable to ask if it is Standardized?

    – Some Name
    Nov 28 '18 at 5:23











  • @SomeName it was, 3 decades ago and has not been for 2 decades.

    – Antti Haapala
    Nov 28 '18 at 5:25











  • @AnttiHaapala So now the behavior is undefined or implementation defined?

    – Some Name
    Nov 28 '18 at 5:27






  • 2





    @SomeName neither. A C99, C11, C17 compiler must diagnose this as an incorrect program.

    – Antti Haapala
    Nov 28 '18 at 5:28













  • @SomeName The support itself for implicit int is removed, though it is still provided as compiler extension. Use strict mode , and your code should fail to compile in first place.

    – Sourav Ghosh
    Nov 28 '18 at 5:29



















So reasonable to ask if it is Standardized?

– Some Name
Nov 28 '18 at 5:23





So reasonable to ask if it is Standardized?

– Some Name
Nov 28 '18 at 5:23













@SomeName it was, 3 decades ago and has not been for 2 decades.

– Antti Haapala
Nov 28 '18 at 5:25





@SomeName it was, 3 decades ago and has not been for 2 decades.

– Antti Haapala
Nov 28 '18 at 5:25













@AnttiHaapala So now the behavior is undefined or implementation defined?

– Some Name
Nov 28 '18 at 5:27





@AnttiHaapala So now the behavior is undefined or implementation defined?

– Some Name
Nov 28 '18 at 5:27




2




2





@SomeName neither. A C99, C11, C17 compiler must diagnose this as an incorrect program.

– Antti Haapala
Nov 28 '18 at 5:28







@SomeName neither. A C99, C11, C17 compiler must diagnose this as an incorrect program.

– Antti Haapala
Nov 28 '18 at 5:28















@SomeName The support itself for implicit int is removed, though it is still provided as compiler extension. Use strict mode , and your code should fail to compile in first place.

– Sourav Ghosh
Nov 28 '18 at 5:29







@SomeName The support itself for implicit int is removed, though it is still provided as compiler extension. Use strict mode , and your code should fail to compile in first place.

– Sourav Ghosh
Nov 28 '18 at 5:29















3














It defaults to an int.



The compiler warning shows what is going on: 



#1 with x86-64 gcc 8.2

<source>:1:9: warning: type defaults to 'int' in declaration of 't' [-Wimplicit-int]

 typedef t;


From C99 onwards, the implicit int rule was removed. So this is not applicable from C99 onward.



If you use the -pedantic-errors compiler option in GCC (meaning strict conformity to the standard), it issues an error. See here.



If you are interested, the relevant section in C89 standard which allowed this:




3.5.2 Type specifiers

Each list of type specifiers shall be one of the following sets; the type specifiers may occur in any order, possibly intermixed with the other declaration specifiers.




  • void

  • char

  • signed char

  • unsigned char

  • short , signed short , short int , or signed short int

  • unsigned short , or unsigned short int

  • int , signed , signed int , or no type specifiers




So in C99, the last part of what was bolded above (or no type specifiers) was removed.






share|improve this answer





















  • 1





    Please don't confuse readers by quoting now-outdated C89, not everyone may know about the applicability of the standards. :)

    – Sourav Ghosh
    Nov 28 '18 at 5:39













  • I did mention that it is not relevant from C99 onwards. But I can make it extra clear.

    – P.W
    Nov 28 '18 at 5:40











  • I would really appreciate if you move the positioning of the two blocks. First, mention about C99/C11 and then, about C89. Just my two cents.

    – Sourav Ghosh
    Nov 28 '18 at 5:45











  • @SouravGhosh: Done.

    – P.W
    Nov 28 '18 at 5:48
















3














It defaults to an int.



The compiler warning shows what is going on: 



#1 with x86-64 gcc 8.2

<source>:1:9: warning: type defaults to 'int' in declaration of 't' [-Wimplicit-int]

 typedef t;


From C99 onwards, the implicit int rule was removed. So this is not applicable from C99 onward.



If you use the -pedantic-errors compiler option in GCC (meaning strict conformity to the standard), it issues an error. See here.



If you are interested, the relevant section in C89 standard which allowed this:




3.5.2 Type specifiers

Each list of type specifiers shall be one of the following sets; the type specifiers may occur in any order, possibly intermixed with the other declaration specifiers.




  • void

  • char

  • signed char

  • unsigned char

  • short , signed short , short int , or signed short int

  • unsigned short , or unsigned short int

  • int , signed , signed int , or no type specifiers




So in C99, the last part of what was bolded above (or no type specifiers) was removed.






share|improve this answer





















  • 1





    Please don't confuse readers by quoting now-outdated C89, not everyone may know about the applicability of the standards. :)

    – Sourav Ghosh
    Nov 28 '18 at 5:39













  • I did mention that it is not relevant from C99 onwards. But I can make it extra clear.

    – P.W
    Nov 28 '18 at 5:40











  • I would really appreciate if you move the positioning of the two blocks. First, mention about C99/C11 and then, about C89. Just my two cents.

    – Sourav Ghosh
    Nov 28 '18 at 5:45











  • @SouravGhosh: Done.

    – P.W
    Nov 28 '18 at 5:48














3












3








3







It defaults to an int.



The compiler warning shows what is going on: 



#1 with x86-64 gcc 8.2

<source>:1:9: warning: type defaults to 'int' in declaration of 't' [-Wimplicit-int]

 typedef t;


From C99 onwards, the implicit int rule was removed. So this is not applicable from C99 onward.



If you use the -pedantic-errors compiler option in GCC (meaning strict conformity to the standard), it issues an error. See here.



If you are interested, the relevant section in C89 standard which allowed this:




3.5.2 Type specifiers

Each list of type specifiers shall be one of the following sets; the type specifiers may occur in any order, possibly intermixed with the other declaration specifiers.




  • void

  • char

  • signed char

  • unsigned char

  • short , signed short , short int , or signed short int

  • unsigned short , or unsigned short int

  • int , signed , signed int , or no type specifiers




So in C99, the last part of what was bolded above (or no type specifiers) was removed.






share|improve this answer















It defaults to an int.



The compiler warning shows what is going on: 



#1 with x86-64 gcc 8.2

<source>:1:9: warning: type defaults to 'int' in declaration of 't' [-Wimplicit-int]

 typedef t;


From C99 onwards, the implicit int rule was removed. So this is not applicable from C99 onward.



If you use the -pedantic-errors compiler option in GCC (meaning strict conformity to the standard), it issues an error. See here.



If you are interested, the relevant section in C89 standard which allowed this:




3.5.2 Type specifiers

Each list of type specifiers shall be one of the following sets; the type specifiers may occur in any order, possibly intermixed with the other declaration specifiers.




  • void

  • char

  • signed char

  • unsigned char

  • short , signed short , short int , or signed short int

  • unsigned short , or unsigned short int

  • int , signed , signed int , or no type specifiers




So in C99, the last part of what was bolded above (or no type specifiers) was removed.







share|improve this answer














share|improve this answer



share|improve this answer








edited Nov 28 '18 at 5:48

























answered Nov 28 '18 at 5:23









P.WP.W

16.7k41455




16.7k41455








  • 1





    Please don't confuse readers by quoting now-outdated C89, not everyone may know about the applicability of the standards. :)

    – Sourav Ghosh
    Nov 28 '18 at 5:39













  • I did mention that it is not relevant from C99 onwards. But I can make it extra clear.

    – P.W
    Nov 28 '18 at 5:40











  • I would really appreciate if you move the positioning of the two blocks. First, mention about C99/C11 and then, about C89. Just my two cents.

    – Sourav Ghosh
    Nov 28 '18 at 5:45











  • @SouravGhosh: Done.

    – P.W
    Nov 28 '18 at 5:48














  • 1





    Please don't confuse readers by quoting now-outdated C89, not everyone may know about the applicability of the standards. :)

    – Sourav Ghosh
    Nov 28 '18 at 5:39













  • I did mention that it is not relevant from C99 onwards. But I can make it extra clear.

    – P.W
    Nov 28 '18 at 5:40











  • I would really appreciate if you move the positioning of the two blocks. First, mention about C99/C11 and then, about C89. Just my two cents.

    – Sourav Ghosh
    Nov 28 '18 at 5:45











  • @SouravGhosh: Done.

    – P.W
    Nov 28 '18 at 5:48








1




1





Please don't confuse readers by quoting now-outdated C89, not everyone may know about the applicability of the standards. :)

– Sourav Ghosh
Nov 28 '18 at 5:39







Please don't confuse readers by quoting now-outdated C89, not everyone may know about the applicability of the standards. :)

– Sourav Ghosh
Nov 28 '18 at 5:39















I did mention that it is not relevant from C99 onwards. But I can make it extra clear.

– P.W
Nov 28 '18 at 5:40





I did mention that it is not relevant from C99 onwards. But I can make it extra clear.

– P.W
Nov 28 '18 at 5:40













I would really appreciate if you move the positioning of the two blocks. First, mention about C99/C11 and then, about C89. Just my two cents.

– Sourav Ghosh
Nov 28 '18 at 5:45





I would really appreciate if you move the positioning of the two blocks. First, mention about C99/C11 and then, about C89. Just my two cents.

– Sourav Ghosh
Nov 28 '18 at 5:45













@SouravGhosh: Done.

– P.W
Nov 28 '18 at 5:48





@SouravGhosh: Done.

– P.W
Nov 28 '18 at 5:48











0














A typedef declaration defines a synonym for an object or pointer type. So you should specify both the type for which you want to create a synonym, and a name to use as synonym.



For example:



// 'byte_t' is a synonym for 'unsigned char'

typedef unsigned char byte_t;

// 'handler_t' is a synonym for 'void (*)(int)', a function pointer

typedef void (*handler_t)(int);

// 'short_p' is a synonym for 'short *'

typedef short * short_p;

// 'record' is a synonym for an anonymous structure

typedef struct {

    int a;

    char *b;

    } record;

// 'error_p' is a synonym for a pointer to 'struct error', defined somewhere else

typedef struct error *error_p;


Many more examples in the source you cited.






share|improve this answer





















  • 1





    This is fine, but how does this answers the question?

    – Sourav Ghosh
    Nov 28 '18 at 5:55











  • @Sourav Ghosh I felt that the typedef syntax was new to him, so I give him some basic examples with comments. Other answers already addressed the compilation problem in itself.

    – A. Semat
    Nov 28 '18 at 6:07











  • Please don't invent your own terms and grammar that don't even exist in the C language. There are no syntax items called "base-type" or "type-alias".

    – Lundin
    Nov 28 '18 at 9:53













  • @Lundin. Thanks. I should have specified that base-type and type-alias should have been replaced by actual types and names. Moreover that pseudo-syntax didn't suit function types. I also changed alias to synonym, since the C standards refers to typedef names as such.

    – A. Semat
    Nov 28 '18 at 10:29
















0














A typedef declaration defines a synonym for an object or pointer type. So you should specify both the type for which you want to create a synonym, and a name to use as synonym.



For example:



// 'byte_t' is a synonym for 'unsigned char'

typedef unsigned char byte_t;

// 'handler_t' is a synonym for 'void (*)(int)', a function pointer

typedef void (*handler_t)(int);

// 'short_p' is a synonym for 'short *'

typedef short * short_p;

// 'record' is a synonym for an anonymous structure

typedef struct {

    int a;

    char *b;

    } record;

// 'error_p' is a synonym for a pointer to 'struct error', defined somewhere else

typedef struct error *error_p;


Many more examples in the source you cited.






share|improve this answer





















  • 1





    This is fine, but how does this answers the question?

    – Sourav Ghosh
    Nov 28 '18 at 5:55











  • @Sourav Ghosh I felt that the typedef syntax was new to him, so I give him some basic examples with comments. Other answers already addressed the compilation problem in itself.

    – A. Semat
    Nov 28 '18 at 6:07











  • Please don't invent your own terms and grammar that don't even exist in the C language. There are no syntax items called "base-type" or "type-alias".

    – Lundin
    Nov 28 '18 at 9:53













  • @Lundin. Thanks. I should have specified that base-type and type-alias should have been replaced by actual types and names. Moreover that pseudo-syntax didn't suit function types. I also changed alias to synonym, since the C standards refers to typedef names as such.

    – A. Semat
    Nov 28 '18 at 10:29














0












0








0







A typedef declaration defines a synonym for an object or pointer type. So you should specify both the type for which you want to create a synonym, and a name to use as synonym.



For example:



// 'byte_t' is a synonym for 'unsigned char'

typedef unsigned char byte_t;

// 'handler_t' is a synonym for 'void (*)(int)', a function pointer

typedef void (*handler_t)(int);

// 'short_p' is a synonym for 'short *'

typedef short * short_p;

// 'record' is a synonym for an anonymous structure

typedef struct {

    int a;

    char *b;

    } record;

// 'error_p' is a synonym for a pointer to 'struct error', defined somewhere else

typedef struct error *error_p;


Many more examples in the source you cited.






share|improve this answer















A typedef declaration defines a synonym for an object or pointer type. So you should specify both the type for which you want to create a synonym, and a name to use as synonym.



For example:



// 'byte_t' is a synonym for 'unsigned char'

typedef unsigned char byte_t;

// 'handler_t' is a synonym for 'void (*)(int)', a function pointer

typedef void (*handler_t)(int);

// 'short_p' is a synonym for 'short *'

typedef short * short_p;

// 'record' is a synonym for an anonymous structure

typedef struct {

    int a;

    char *b;

    } record;

// 'error_p' is a synonym for a pointer to 'struct error', defined somewhere else

typedef struct error *error_p;


Many more examples in the source you cited.







share|improve this answer














share|improve this answer



share|improve this answer








edited Nov 29 '18 at 20:38

























answered Nov 28 '18 at 5:51









A. SematA. Semat

584




584








  • 1





    This is fine, but how does this answers the question?

    – Sourav Ghosh
    Nov 28 '18 at 5:55











  • @Sourav Ghosh I felt that the typedef syntax was new to him, so I give him some basic examples with comments. Other answers already addressed the compilation problem in itself.

    – A. Semat
    Nov 28 '18 at 6:07











  • Please don't invent your own terms and grammar that don't even exist in the C language. There are no syntax items called "base-type" or "type-alias".

    – Lundin
    Nov 28 '18 at 9:53













  • @Lundin. Thanks. I should have specified that base-type and type-alias should have been replaced by actual types and names. Moreover that pseudo-syntax didn't suit function types. I also changed alias to synonym, since the C standards refers to typedef names as such.

    – A. Semat
    Nov 28 '18 at 10:29














  • 1





    This is fine, but how does this answers the question?

    – Sourav Ghosh
    Nov 28 '18 at 5:55











  • @Sourav Ghosh I felt that the typedef syntax was new to him, so I give him some basic examples with comments. Other answers already addressed the compilation problem in itself.

    – A. Semat
    Nov 28 '18 at 6:07











  • Please don't invent your own terms and grammar that don't even exist in the C language. There are no syntax items called "base-type" or "type-alias".

    – Lundin
    Nov 28 '18 at 9:53













  • @Lundin. Thanks. I should have specified that base-type and type-alias should have been replaced by actual types and names. Moreover that pseudo-syntax didn't suit function types. I also changed alias to synonym, since the C standards refers to typedef names as such.

    – A. Semat
    Nov 28 '18 at 10:29








1




1





This is fine, but how does this answers the question?

– Sourav Ghosh
Nov 28 '18 at 5:55





This is fine, but how does this answers the question?

– Sourav Ghosh
Nov 28 '18 at 5:55













@Sourav Ghosh I felt that the typedef syntax was new to him, so I give him some basic examples with comments. Other answers already addressed the compilation problem in itself.

– A. Semat
Nov 28 '18 at 6:07





@Sourav Ghosh I felt that the typedef syntax was new to him, so I give him some basic examples with comments. Other answers already addressed the compilation problem in itself.

– A. Semat
Nov 28 '18 at 6:07













Please don't invent your own terms and grammar that don't even exist in the C language. There are no syntax items called "base-type" or "type-alias".

– Lundin
Nov 28 '18 at 9:53







Please don't invent your own terms and grammar that don't even exist in the C language. There are no syntax items called "base-type" or "type-alias".

– Lundin
Nov 28 '18 at 9:53















@Lundin. Thanks. I should have specified that base-type and type-alias should have been replaced by actual types and names. Moreover that pseudo-syntax didn't suit function types. I also changed alias to synonym, since the C standards refers to typedef names as such.

– A. Semat
Nov 28 '18 at 10:29





@Lundin. Thanks. I should have specified that base-type and type-alias should have been replaced by actual types and names. Moreover that pseudo-syntax didn't suit function types. I also changed alias to synonym, since the C standards refers to typedef names as such.

– A. Semat
Nov 28 '18 at 10:29


















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