6 replies to this topic

[TheUmer]

I'm using the following code to do so:

#include <stdio.h>

int main ()
{
int i;
char ch[20]="1 2 34 4 56672332432";
int ch1[20];
for (i=0; i<20; i++)
{
      ch1[i]=(int)ch[i]-'0';
}
for (i=0; i<20; i++)
{
      printf ("%d\n ", ch1[i]);
}
return 0;
}

The problem is I can't an integer more than 9 as it is. For example, the array is reading and converting each char separately, so for example if I want to get 34 form char array into an integer 34, it won't do so. It picks and gives 3 separately and 4 separately.

What should I do for this problem?

[Moudi]

example
char myarray[] = "1,2,3,4,5"
int array[5]
for(int i=0; i<5; i++)
{
array[i] = stoi(myarray[i]);
}

[TheUmer]

#include <stdlib.h> this would be used for using stoi, right? But compiler is giving an error: "undefined reference to stoi"

What to be done?

[Moudi]

I'm sorry its atoi

fix it xD

[TheUmer]

Fine. But when used atoi instread of stoi, it gives the following error:

"passing argument 1 of ‘atoi’ makes pointer from integer without a cast."

What now? :s

[Moudi]

Uhm, sorry i can't get it to work :/

[fread]

I think you should rethink your strategy. A char array will count that whitespace as a valid char and any manipulation you make on the items in the array will be done on the whitespace char. Dont forget a char is a single character. I think things like 34 would be counted as two chars since each char occupies one position in the array and 34 would occupy two.