Rank: ? (6)
Member #: 28706

Hi,

a small malloc problem .

Below is part of the code of a function for converting float values into floating point:

Code:

char* floating_point(float flt)
{
char *formula;

formula = malloc(21);

return formula;
}
The program exits without warning when it reaches the malloc line in the code above.

To be complete, here is how I call the function:

Code:
fa_string = floating_point(formula_a);

fprintf(fpout,"%s\n", fa_string);

free(fa_string);

What is wrong here?

Thanks for any help.

Rank: ? (119)
Member #: 28292

The entire code please

Rank: ? (6)
Member #: 28706

Hi igorok,

Here is the full code...

/******************************************************************************/
char* floating_point(float flt)
{
char hex[9], hex_2[9], temp[3], *p;
char *formula;
int decimal[4], i=0, ii=0;
short cnt=0, cnt2=0;

union
{
unsigned long i;
float f;
}v;

if(!(formula = malloc(21)))
{
free(formula);//free up resources
if(!(formula = malloc(21)))
exit(-1);
}

v.f = flt;
sprintf(hex_2,"%08lx",v.i);
puts(hex_2);

cnt=7;
while(cnt2<=3)
{

sprintf(temp,"%c%c",hex_2[cnt-1],hex_2[cnt]);
printf("%s\t",temp);
decimal[cnt2] = strtol(temp,&p,16);
cnt-=2;
cnt2++;
}
printf("\n";
sprintf(formula,"%d\t%d\t%d\t%d",decimal[0],decimal[1],decimal[2],decimal[3]);
puts(formula);

return formula;
}
/******************************************************************************/

Thanks

Rank: ? (4)
Member #: 28871

Replace all `malloc(21);` calls with `(char*)malloc(21);` and you're set.

Cheers,
T-Metal