How do I concatenate an integer onto a string in C?

Question

The part of the code in question is attempting to decode what register is being used by a MIPS instruction.

This does it by passing in an integer value of the register and then should return a string containing the name of the register. The prince statement that does it is here, where it calls getReg to get the string.

printf("$%d aka $%s\n", itype->rs, getReg(itype->rs));

So far, I've tried this to concatenate them (without the case statements):

char* getReg(int d) { char out[4]; sprintf(out, "a%d", (d - 4)); return out; }

But the output results in this:

$6 aka $ìü(

When it should be:

$6 aka $a2

Where am I going wrong with this?

Whether out is large enough is irrelevant, returning a local array is still undefined behavior. — Anton Kovalenko
– Anton Kovalenko, Commented Feb 12, 2013 at 16:00
True, try to give a printf after your sprintf and see. Also, use the GDB, helps a lot — noMAD
– noMAD, Commented Feb 12, 2013 at 16:01

Grijesh Chauhan · Accepted Answer · 2013-04-02 19:00:38Z

You are returning address of a local variable (out).

char* getReg(int d) { char out[4]; sprintf(out, "a%d", (d - 4)); return out; }

scope and life of out is within function getReg() only.

Allocate memory dynamically for out to return and access outside function. (and large enough), like below

#define SIZE 25 char* getReg(int d) { char *out = malloc(SIZE*sizeof(char)); sprintf(out, "a%d", (d - 4)); // I don't know about calculation?? return out; }

and don't forget to free memory.

@H2CO3 Thanks for char* out. And yes sizeof(char) is 1 but I always do this way :)

John Bode · Accepted Answer · 2013-02-12 16:04:47Z

The array out is local to your getreg function; once the function exits, out no longer exists and the pointer value you return is no longer valid.

It's better to pass the output array as a parameter to the function:

void getReg(int d, char *str) { sprintf(str, "a%d", (d-4)); }

and call it as

char mystr[4]; getReg(r, mystr);

alk · Accepted Answer · 2013-02-12 17:54:01Z

As already mentioned by others the OP returns a reference to storage, which is already invalid if used in the call to printf().

An alternative to provide an external buffer whould be this:

char * getReg(int d, char * out) { sprintf(out, "a%d", (d - 4)); return out; } ... printf( "$%d aka $%s\n", itype->rs, getReg( itype->rs, (char[32]){0} /* provide a 32 byte long buffer, initialised to 0s */ ) );

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