Thread: Struct and typedef

Hi, I have this code and I wants by using struct to insert "1" to the size and afterwards print it, but it gives a "syntax ; error" although all ";" decent in my code's script.
code:

Code:

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


typedef struct
{
	char num[4];
	char ters[4];
	int  size;
	char *address;
	 


}memIdentifier;


main()
{
	memIdentifier.size=1;
        memIdentifier.addres='Hello';
	printf("%d", memIdentifier.size);
        printf("%c", memIdentifier.address);
	return 0;
}

any help for how to correct it for printing? thanks.

Originally Posted by laserlight

The typedef means that memIdentifier is a type, not a variable name. You need to declare a variable.

GOT you, but got wrong printing-it prints 1 but in the sl.address prints something weird!(not like input at all)
Here's the code after edition:
code:

Code:

typedef struct
{
    char num[4];
    char ters[4];
    int  size;
    char *address;
      
 
 
}memIdentifier;
 
 
main()
{
	memIdentifier sl;
    sl.size=1;
	sl.address="hello";
    printf("%d", sl.size);
	printf("%c", sl.address);
    return 0;
}

Moreover, I have another question, when I write in the struct's structure itself a pointer like "char*address" why if I would call it I write the variable.address, and not variable.(*address) ?
thanks

Where have you allocated memory for that pointer in your structure?

You really really need to start finding and reading the documentation for the standard functions before you try to use them. In this instance you need to find and read the documentation for printf() paying very close attention to the format specifiers. Much like scanf() you must insure that the format specifiers match the data types.

Jim

Originally Posted by jimblumberg

Where have you allocated memory for that pointer in your structure?

You really really need to start finding and reading the documentation for the standard functions before you try to use them. In this instance you need to find and read the documentation for printf() paying very close attention to the format specifiers. Much like scanf() you must insure that the format specifiers match the data types.

Jim

You are always saying that and the same content, I already have read them and dont make from your self something you don't really know about! I'm sorry for saying that but from your question "where have you allocated...." sounds you are really still not skilled of, and just throwing silly tips arbitrary.

You always saying that and the same content, I already have read them and dont make from your self something you don't really know about!

Then you need to re-read the documentation, since you still don't seem to understand how to use the function. Why are you trying to use the specifier for a single character for a value that you seem to think is a string?

I'm sorry for saying that but from your question "where have you allocated...." sounds you are really still not skilled of, and just throwing silly tips arbitrary.

Well then why can't you answer the question? The answer of course if you never allocated any memory for that pointer.

And as I said in your last topic, I don't know what compiler you're using but you are either ignoring warnings and errors or your compiler is defective. Your code has several deficiencies that a decent compiler should be able to warn you about. Look at what my compiler says about your code.

main.c|13|error: return type defaults to ‘int’|
main.c||In function ‘main’:|
main.c|18|error: implicit declaration of function ‘printf’ [-Wimplicit-function-declaration]|
main.c|18|warning: incompatible implicit declaration of built-in function ‘printf’|
main.c|19|warning: format ‘%c’ expects argument of type ‘int’, but argument 2 has type ‘char *’ [-Wformat=]|
||=== Build finished: 2 errors, 2 warnings (0 minutes, 0 seconds) ===|

Also you should know by now that you can't use the assignment operator to assign values to strings.

Jim

Originally Posted by vart

address is pointer to string - so it should be printed using %s format, %c is for 1 character, not string

Alright, but isn't *address mean like we go to the value that's found in pointer address? and that pointer isn't gonna be changed unless I ordered that.

Originally Posted by jimblumberg

Then you need to re-read the documentation, since you still don't seem to understand how to use the function. Why are you trying to use the specifier for a single character for a value that you seem to think is a string?


Well then why can't you answer the question? The answer of course if you never allocated any memory for that pointer.

And as I said in your last topic, I don't know what compiler you're using but you are either ignoring warnings and errors or your compiler is defective. Your code has several deficiencies that a decent compiler should be able to warn you about. Look at what my compiler says about your code.



Also you should know by now that you can't use the assignment operator to assign values to strings.

Jim

what compiler are you using? that's not appearing for me at all, if it would, then I wouldn't ask at all .