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I'm having some issues with storing an address of a specific value in an array of strings to a pointer and printing it out. Please excuse the bad variable names, they are just for the example.

char **code;                      // code is an array of 100 strings of length 8 characters
code = malloc (100*sizeof(*code));
for (i=0; i<100; i++) {
    code[i]=malloc(8*sizeof(**code));
}

char **r;                        // r is an array of 16 strings of 32 characters
r = malloc (16*sizeof(*r));
for (i = 0; i < 16; i++)
    r[i] = malloc(32*sizeof(**r));

char *a;                         // a is a pointer to a string

a = (char *) &r[13];             // point a to value at r[13]

*a = (char *)&code[100];         // trying to assign the value of r[13] to the address of code[100]   

printf("Code 100 Add: %p a Val: %i\n", &code[100], *sp);  // trying to check values.

I'm trying to assign the value of a (which points to r[13], so assign value of r[13]) to the value of the Address of the string at code[100]. Is even a string of 32 characters the best way to do this?

Appreciate any help, Gareth

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  • This example has a lot of flaws. Examples: not casting the result of malloc, assigning a (char *) to both (a) and (*a). Can you post real code? What is it that you are trying to do? If you are having problems compiling - start with the compiler errors. If you are having problems trying to get it to run - please post code that compiles and start with a good debugger and smaller arrays so that you can see what is going on at each step. Commented Apr 27, 2011 at 17:37

6 Answers 6

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a = (char *) &r[13];             // point a to value at r[13]

Turn on your compiler warnings, and pay attention to what the compiler tell you when you remove this cast. You shouldn't need any casts in this code.

The type of r is char** and so the type of r[13] is char*, and the type of &r[13] is char**, which you're assigning to a char*.

P.s., next time please also include the actual error you receive vs what you expected.

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To assign a string, use strncpy. Don't copy the pointer value directly because you will free it twice later this way (among other problems).

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So something like: strcpy(a, &code[100]); It still doesn't seem to store the correct value according the the printf
You don't need the extra &, because code[100] already points to the beginning of a string.
Safest is strncpy(a, code[100], 31); a[31] = '\0'; You should use a constant instead of 32 literally, of course.
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&r[13] is not a pointer to char.

You just need to 

r[13] = code[100]

I do agree with Frederik that you should be careful when free()-ing the allocated memory since you now have two pointers pointing at the same memory block. If you prefer to follows his advice, try the following:

strncpy(r[13], code[100], 8)
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Rather than assigning a pointer to a memory address, you're going to want to copy the data at that memory address using strcpy.

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With definition of char *a; the *a on the left-hand side of an assignment becomes an lvalue of type char. You can assign pointer values there as integers, as in square peg into a round hole, though it does not make much sense. To copy strings use str[nl]cpy(3).

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I hate to ask a question during someone else's question... but shouldn't the char array setup/malloc calls be something more like this? Isn't he allocating too much with sizeof(**code)? And then... if it's for 8 characters... won't we want 9 to make room for '\0'?

char **code;                      // code is an array of 100 strings of length 8 characters
code = (char**) malloc (100*sizeof(char*));
for (i=0; i<100; i++) {
   code[i] = (char*)malloc(9*sizeof(char));
}
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