int to char array?

[blackeyrichards]

This is (hopefully) just a quicky. How do I convert an int to a character array?

I.E., I want to write a correct/working version of this:

int numI = 34;
char numA[255] = (char[255]) numI;

cout<<numA;

Gives the output:34

Thanks,
Paul

 

[Bangieff]

well I didn't try it but the algorthm should be something like that I think (I'm not very fanous with the C math functions so there should be a better way):

#include <stdio.h>

main()
{
int tmp,numI,i,j;
char *numA;

tmp=numI=12345;
i=j=0;
while (tmp>=1)
{
i++;tmp=(int)tmp/10;
}
numA=(char *)malloc(sizeof(char)*i);
for(j=0;j<i;j++)
{
tmp=(int)numI/10;
tmp=numI-10*tmp;
numI=(int)numI/10;
numA[i-j-1]=(char)tmp+48;
}
printf(&quot;string -> &quot;
for(j=0;j<i;j++)
{
printf(&quot;%c&quot;,numA[j]);
}
printf(&quot;\n&quot;
}

 

[Bangieff]

uups excuse me you can just replace this:

printf(&quot;string -> &quot;
for(j=0;j<i;j++)
{
printf(&quot;%c&quot;,numA[j]);
}
printf(&quot;\n&quot;

with this:
printf(&quot;string -> %s\n&quot;,numA);

sorry about that

 

[Lim]

just one line:
sprintf(numA,&quot;%d&quot;,numI);
that is it.

 

[rbobbitt]

Oops:

sprintf(numA,&quot;%d&quot;,&numI);


Russ
bobbitts@hotmail.com

http://home.earthlink.net/~bobbitts

 

[Bangieff]

huh a little bit shorter than mine