What does typedef with no type mean?

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?

edited Nov 29 '18 at 20:40

dbush

102k13108144

asked Nov 28 '18 at 5:16

Some Name

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

add a comment |

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?

edited Nov 29 '18 at 20:40

dbush

102k13108144

asked Nov 28 '18 at 5:16

Some Name

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

add a comment |

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?

edited Nov 29 '18 at 20:40

dbush

102k13108144

asked Nov 28 '18 at 5:16

Some Name

1,517417

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

edited Nov 29 '18 at 20:40

dbush

102k13108144

asked Nov 28 '18 at 5:16

Some Name

1,517417

edited Nov 29 '18 at 20:40

dbush

102k13108144

asked Nov 28 '18 at 5:16

Some Name

1,517417

edited Nov 29 '18 at 20:40

dbush

102k13108144

edited Nov 29 '18 at 20:40

dbush

102k13108144

edited Nov 29 '18 at 20:40

dbush

102k13108144

asked Nov 28 '18 at 5:16

Some Name

1,517417

asked Nov 28 '18 at 5:16

Some Name

1,517417

asked Nov 28 '18 at 5:16

Some Name

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

add a comment |

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

add a comment |

3 Answers
3

active

oldest

votes

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.

edited Nov 28 '18 at 5:52

answered Nov 28 '18 at 5:22

Sourav Ghosh

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

add a comment |

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.

edited Nov 28 '18 at 5:48

answered Nov 28 '18 at 5:23

P.W

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

add a comment |

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.

edited Nov 29 '18 at 20:38

answered Nov 28 '18 at 5:51

A. Semat

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

add a comment |

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3 Answers
3

active

oldest

votes

3 Answers
3

active

oldest

votes

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.

edited Nov 28 '18 at 5:52

answered Nov 28 '18 at 5:22

Sourav Ghosh

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

add a comment |

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.

edited Nov 28 '18 at 5:52

answered Nov 28 '18 at 5:22

Sourav Ghosh

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

add a comment |

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.

edited Nov 28 '18 at 5:52

answered Nov 28 '18 at 5:22

Sourav Ghosh

111k15131190

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.

edited Nov 28 '18 at 5:52

answered Nov 28 '18 at 5:22

Sourav Ghosh

111k15131190

edited Nov 28 '18 at 5:52

answered Nov 28 '18 at 5:22

Sourav Ghosh

111k15131190

answered Nov 28 '18 at 5:22

Sourav Ghosh

111k15131190

answered Nov 28 '18 at 5:22

Sourav Ghosh

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

add a comment |

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

@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

add a comment |

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.

edited Nov 28 '18 at 5:48

answered Nov 28 '18 at 5:23

P.W

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

add a comment |

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.

edited Nov 28 '18 at 5:48

answered Nov 28 '18 at 5:23

P.W

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

add a comment |

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.

edited Nov 28 '18 at 5:48

answered Nov 28 '18 at 5:23

P.W

16.7k41455

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.

edited Nov 28 '18 at 5:48

answered Nov 28 '18 at 5:23

P.W

16.7k41455

edited Nov 28 '18 at 5:48

answered Nov 28 '18 at 5:23

P.W

16.7k41455

answered Nov 28 '18 at 5:23

P.W

16.7k41455

answered Nov 28 '18 at 5:23

P.W

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

add a comment |

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

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

add a comment |

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.

edited Nov 29 '18 at 20:38

answered Nov 28 '18 at 5:51

A. Semat

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

add a comment |

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.

edited Nov 29 '18 at 20:38

answered Nov 28 '18 at 5:51

A. Semat

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

add a comment |

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.

edited Nov 29 '18 at 20:38

answered Nov 28 '18 at 5:51

A. Semat

584

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.

edited Nov 29 '18 at 20:38

answered Nov 28 '18 at 5:51

A. Semat

584

edited Nov 29 '18 at 20:38

answered Nov 28 '18 at 5:51

A. Semat

584

answered Nov 28 '18 at 5:51

A. Semat

584

answered Nov 28 '18 at 5:51

A. Semat

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

add a comment |

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

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

add a comment |

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