struct initialization failure

Question

I'm having difficulty initializing a struct in code per below. Can this even be done, or do I really need to memcpy (urg) the 5-character string into the struct?

struct MyStruct  
{  
    char x[5];  
};

main(...)  
{  
    const char* MyString = {"mnopq"}; // I understand this is a non-NULL terminated string -  
                                  // it's OK, I just want five character fields in an array

struct MyStruct = {MyString};     // <---This gives warnings below  
}

warning: missing braces around initializer
warning: initialization makes integer from pointer without cast

If I wrap initializer like:

struct MyStruct = {{MyString}};

the first warning goes away. The second warning does not. And, thus, the struct is not initialized as hoped.

Thanks in advance for help.

@user654077: Sorry, but your question and your comments contradict each other. If it should be const char* MyString = "mnopq", then it is a null-terminated string. Yet, in the code comments you say that it is not. ??? Decide what you want and edit your question. At this time it is a mess. — AnT stands with Russia
– AnT stands with Russia, Commented Mar 10, 2011 at 18:37

Erik · Accepted Answer · 2011-03-10 18:42:54Z

4

"mnopq" is an array, const char * MyString is not.

struct MyStruct foo = {"abcde"}; will work, whereas your approach converts the pointer MyString to an integral value and assigns it to the first element of x.

edited Mar 10, 2011 at 18:42

answered Mar 10, 2011 at 18:37

Erik

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Comments

DavidMFrey · Accepted Answer · 2011-03-11 03:36:38Z

3

You are trying to fill a char array with a pointer. MyString is seen as a const char * instead of an array. Change your type in the structure to const char *x, and it should do the trick. Also, MyString is actually a NULL terminated string here. The compiler sets it into memory with the extra NULL byte at the end, then treats MyString like a const char *.

If you do want the char x[5] instead of the const char *, you could initialize in a couple of ways,

struct MyStruct ms = {{'m', 'n', 'o', 'p', 'q'}};

or you could do

struct MyStruct ms = {{ MyString[0], MyString[1], MyString[2], MyString[3], MyString[4]}};

The first set of braces is for initializing the struct members, the second set is for initializing the char array members, so each member(char) has to be set individually.

edited Mar 11, 2011 at 3:36

answered Mar 10, 2011 at 18:29

DavidMFrey

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2 Comments

Rob Over a year ago

I edited above to have an identifier. One other thing is that the MyString[index] version will print junk at the end whereas the first version does not (e.g. mnopq¦ah¦)

DavidMFrey Over a year ago

Right, because there is no NULL terminator. But there shouldn't be a NULL at the end of the first example either... I don't think he mentioned using a defined print function to print out the array as a string.

Cristian Toader · Accepted Answer · 2011-03-10 18:40:45Z

0

You need to declare a variable of the type MyStruct and assign it's x value to that string. This should work:

struct MyStruct {
char x[5];
};

int main(int argc, char *argv[]) {

   struct MyStruct a;
   const char* MyString = "mnop";

   strncpy((a.x), MyString, 5);

   /* 
    *if within the 5 characters you do not have the null char it will 
    * also print garbage
    */
   printf("%s", a.x);

   return 0;
}

However you can't assign a pointer to const char to an array of char as they are different types. So you would need to either use char* in the struct or use memcpy / strcpy to get the information.

answered Mar 10, 2011 at 18:40

Cristian Toader

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Comments

Rob · Accepted Answer · 2011-03-10 19:53:12Z

0

The answer by DavidMFrey is correct and should get credit before mine! However, here's the source code using his answer as I was able to run it:

#include <stdio.h>
struct MyStruct  
{  
    char x[5];  
};
int main(void)  
{  
    const char* MyString = "mnopq";
    struct MyStruct foo = {{ MyString[0], MyString[1], MyString[2], MyString[3], MyString[4]}};
    printf("foo.x is: %s\n", foo.x); // outputs: foo.x is: mnopq¦ah¦"

    struct MyStruct ms = {{'m', 'n', 'o', 'p', 'q'}};
    printf("ms.x is: %s\n", ms.x);   // ms.x is: mnopq

    return 0;
}

answered Mar 10, 2011 at 19:53

Rob

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