typedef required in struct declaration

Question

I'm trying to create an array of struct elements, as shown below:

#include <stdio.h>
#include <stdlib.h>

struct termstr{
double coeff;
double exp;
};

int main(){

termstr* lptr = malloc(sizeof(termstr)*5);

return 0;
}

When i compile this, i get errors as follows:

term.c: In function ‘main’:
term.c:11:1: error: unknown type name ‘termstr’
term.c:11:31: error: ‘termstr’ undeclared (first use in this function)

However, when i change my code to the following, it compiles as usual:

#include <stdio.h>
#include <stdlib.h>

typedef struct termstr{
double coeff;
double exp;
}term;

int main(){

term* lptr = malloc(sizeof(term)*5);

return 0;
}

I've added typedef (with type name as term), changed the name of struct to termstr and am allocating memory with term* as the type of pointer.

Is typedef always required for such a situation i.e. for creating arrays of structs? If not, why was the first code giving errors? Is typedef also required to create and use a single instance of a struct?

possible duplicate of What's the syntactically proper way to declare a C struct? — Oliver Charlesworth
– Oliver Charlesworth, Commented Jun 15, 2013 at 14:57
No, you dont need typedef. You could also do without, but you'll need to add struct keyword: struct termstr *lptr = malloc(5 * sizeof *lptr); — wildplasser
– wildplasser, Commented Jun 15, 2013 at 14:58

Community · Accepted Answer · 2017-05-23 12:13:00Z

2

First type is not working because you have forgot struct keyword before termstr. Your data type is struct termstr but not just termstr. When you typedef, the resulting name is used as an alias for struct termstr.

Even you don't need to do that. Using typedef is better:

By the way don't forget to free the memory:

read why to use typedef?

Your working code should be:

#include <stdio.h>
#include <stdlib.h>

struct termstr{
  double coeff;
  double exp;
};

int main(){

struct termstr* lptr = malloc(sizeof(struct termstr)*5);
free(lptr);
return 0;
}

edited May 23, 2017 at 12:13

CommunityBot

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answered Jun 15, 2013 at 15:00

pinkpanther

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Comments

alk · Accepted Answer · 2013-06-15 14:56:33Z

1

It should be:

struct termstr * lptr = malloc(sizeof(struct termstr)*5);

or even better:

struct termstr * lptr = malloc(sizeof(*lptr)*5);

answered Jun 15, 2013 at 14:56

alk

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Comments

Thomas Padron-McCarthy · Accepted Answer · 2013-06-15 14:57:44Z

1

In C, the name of the data type is "struct termstr", not just "termstr".

answered Jun 15, 2013 at 14:57

Thomas Padron-McCarthy

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Comments

Maroun · Accepted Answer · 2013-06-15 15:06:33Z

1

You can do something like this:

typedef struct termstr{
   double coeff;
   double exp;
} termstrStruct;

And then you can use only termstrStruct as the struct's name:

termstrStruct* lptr = malloc(sizeof(termstrStruct)*5);

It is not always required, you can simply write struct termstr.

Don't forget to free the allocated memory!

edited Jun 15, 2013 at 15:06

answered Jun 15, 2013 at 14:58

Maroun

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2 Comments

alk Over a year ago

Although this is perfectly valid, I'd consider it code obfuscation naming different things the same.

Maroun Over a year ago

Agreed. I don't know exactly what did he mean by this struct, but I edited it to have a different name.

jh314 · Accepted Answer · 2013-06-15 15:11:31Z

1

Typedef is a convenient way of shortening this:

struct termstr* lptr = (struct termstr*)malloc(sizeof(struct termstr)*5);

to this:

typedef struct termstr* term;
term* lptr = (term*)malloc(sizeof(term)*5);

Casting the malloc is also a good idea!

edited Jun 15, 2013 at 15:11

answered Jun 15, 2013 at 15:00

jh314

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4 Comments

wildplasser Over a year ago

Casting the malloc is a terrible idea. IMO. (BTW: you'd need an extra asterix there)

alk Over a year ago

Why casting malloc() in C is unnecessary and even more a bad idea: stackoverflow.com/a/605858/694576

jh314 Over a year ago

Huh, my profs always said casting was better style. Well, you learn something new everyday!

Alex Celeste Over a year ago

General rule: when you have the choice to cast or not, casting is poor style. Casting means telling the compiler "no, you're wrong, I know better". If you don't have to tell the compiler this, don't; the compiler's job is to help you with its complaints.

urzeit · Accepted Answer · 2013-06-15 14:59:07Z

0

If you want to use the typename termstr on it's own you can use typedef: typedef struct { double a; double b; } termstr;

answered Jun 15, 2013 at 14:59

urzeit

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Comments

amo-ej1 · Accepted Answer · 2013-06-15 14:59:11Z

0

In C you need to add the struct keyword as well, so either you use a typedef to link an alias with 'struct termstr' or you need to write something like

struct termstr* lptr = malloc(sizeof(struct termstr)*5);

In C++ however you can reference it as 'termstr' directly (read: the struct keyword isn't required there anymore).

answered Jun 15, 2013 at 14:59

amo-ej1

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