如何定义指向结构的指针

Question

I know this is a very basic problem, but I cannot move forward without it and its not clearly explained elsewhere.

Why is this programming giving me so many errors of undeclared identifier? I have declared it, though.

These are the error i am getting.

Error   2   error C2143: syntax error : missing ';' before 'type'
Error   3   error C2065: 'ptr' : undeclared identifier
Error   4   error C2065: 'contactInfo' : undeclared identifier
Error   5   error C2059: syntax error : ')'
Error   15  error C2223: left of '->number' must point to struct/union

and more...

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

typedef struct contactInfo
{
    int number;
    char id;
}ContactInfo;


void main()
{

    char ch;
    printf("Do you want to dynamically etc");
    scanf("%c",&ch);
    fflush(stdin);


        struct contactInfo nom,*ptr;
        ptr=(contactInfo*)malloc(2*sizeof(contactInfo));

    nom.id='c';
    nom.number=12;
    ptr->id=nom.id;
    ptr->number=nom.number;
    printf("Number -> %d\n ID -> %c\n",ptr->number,ptr->id);

}

Answer 1

typedef struct contactInfo
{
    int number;
    char id;
}ContactInfo;

This code defines 2 things:

1. a type named ContactInfo
2. a struct named contactInfo

Notice the difference of the c and C !

In your code you are using a mixed combination of both, which is allowed (although confusing IMHO).

If you use the struct variant you need to explicitly use struct contactInfo . For the other variant ( ContactInfo ) you must to omit the struct part as it is part of the type definition alteady.

So be careful with both different definitions of your structure. Best would be to only use either one of the variants.

---

I do not have Visual Studio at hand, but the following (corrected) code compiles with gcc properly without any warnings:

#include<stdlib.h>

typedef struct contactInfo
{
    int number;
    char id;
}ContactInfo;


void main()
{
    ContactInfo nom,*ptr;
    ptr=malloc(2*sizeof(ContactInfo));    
}

(I left out the not so interesting/unmodified parts of the code)

Answer 2

This:

ptr=(contactInfo*)malloc(2*sizeof(contactInfo));

is wrong, there's no type called contactInfo .

There's a struct contactInfo , which is typedef :d as ContactInfo . C is case-sensitive (and you must include the struct unlike how it works in C++).

Also note that the quoted line is better written as:

ptr = malloc(2 * sizeof *ptr);

Since casting the return value of malloc() is a bad idea in C , I removed the cast. Also, repeating the type is a bad idea since it makes the risk of introducing error greater, so I removed that as well.

Note that sizeof *ptr means "the size of the the type that ptr points at" and is a very handy thing.

Answer 3

Replace

ptr=(contactInfo*)malloc(2*sizeof(contactInfo));

with

ptr=malloc(2*sizeof(struct contactInfo));

Answer 4

C is a case sensitive language.

ptr=(contactInfo)malloc(2*sizeof(contactInfo));

which should be:

ptr=malloc(2*sizeof(ContactInfo));

Answer 5

    struct contactInfo nom,*ptr;
    ptr=(contactInfo*)malloc(2*sizeof(contactInfo));

Since you have defined a typedef, the first statement above can be written as below, though what you have written is also correct but that does not effectively use the typedef you defined.

  ContactInfo nom,*ptr;

Also, since C is case sensitive language,use as you defined in typedef. Capitalise C in contactinfo

The typecasting you are doing should not be done and considered to be a bad practise. As void * is automatically and safely promoted to any other pointer type in this case of malloc .

ptr=malloc(2*sizeof(ContactInfo));

Answer 6

struct contactInfo nom,*ptr;
ptr=(contactInfo*)malloc(2*sizeof(contactInfo));

here you are using contactInfo to typcast where as it should be struct contactInfo. and as you have typedef it into ContactInfo you can use that also.

    struct contactInfo nom,*ptr;
    ptr=(ContactInfo*)malloc(2*sizeof(ContactInfo));