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

4

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.

edited Apr 2, 2013 at 19:00

answered Feb 12, 2013 at 16:01

Grijesh Chauhan

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2 Comments

user529758 Over a year ago

1. it should be char *out, 2. sizeof(char) is 1.

Grijesh Chauhan Over a year ago

@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

0

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is local to your getreg function; once the function exits, out no longer exists and the pointer value you return is no longer valid.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

answered Feb 12, 2013 at 16:04

John Bode

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Comments

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

0

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 */
  )
);

edited Feb 12, 2013 at 17:54

answered Feb 12, 2013 at 17:48

alk

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