atoi & scanf

Code:

---

#include <stdio.h>
#include <stdlib.h>


int main(void)
{
        char input[100];
        int z;


        printf("Insert string: ");
        scanf("%s", input);


        z = atoi(input);


        printf("%d", z);


        return 0;
}

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Please, z is returning zero. Would love an explanation with a fix.

atoi(3) - Linux manual page

Tim S.

Quote:

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Originally Posted by stahta01

atoi(3) - Linux manual page

Tim S.

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T

This actually isn't what I want. I wanted to write a program to convert CHAR into ASCII code.

I suggest you post the input with the output you want because what you stated is not clear to me.
To me char is already ASCII!

Tim S.

Perhaps you want to print the value of each character?

Code:

---

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

int main(void)
{
    char input[100];

    printf("Insert string: ");
    scanf("%99s", input);

    for(size_t i = 0; i < strlen(input); ++i)
        printf("%c = 0x%X = %d\n", input[i], (int)input[i], (int)input[i]);

    return 0;
}

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Quote:

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convert CHAR into ASCII

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Code:

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#include <stdio.h>
#include <stdlib.h>
#include <string.h>

int main( int argc, char *argv[] ) {
   
    char input[100];

    printf( "String: " ); scanf( "%s", input );

    for (int i=0; i<strlen(input); i++) {

        int z = (int)(input[i]);

        printf( "%i,", z );

    }
 
    printf( "\n" ); 
    return 0;
}

---

C - Type Casting - Tutorialspoint

Quote:

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atoi & scanf

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Code:

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#include <stdio.h>
#include <stdlib.h>

int main( int argc, char *argv[] ) {
   
  char number[128];

  printf( "Enter Number: ");
  scanf( "%s", number);
 
  int z = atoi( number );
  printf( "%i \n", z );

  return 0;
}

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Looks good, thanks to all those that helped! Great to be back on the forum!