将字符串转换为C中的数字

Question

I'm trying to write a program which converts lower cased alphabet to numerical digits. a -> 01 b -> 02 ... z -> 26

For the first nine letters I need to put a 0 before the number.

This is my code.

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


#define MAXLEN 128

void s2n(char str[])
{

  int i;
  char x;
  char result[256];

  for (i=0; str[i] != '\0'; i++) {
    x = str[i];
    x = x - 96;


    if (x < 10) {

      char src[2] = {0, x};
      strcat(result, src);
    }
    else {
      char src2[1] = {x};
      strcat(result, src2);
    }

    printf("%s",  result);

  }

}

int main(void)
{
  char str[MAXLEN];

  printf("Lower cased string please: ", MAXLEN);
  scanf("%s", str);

  s2n(str);

  return 0;

}

Could you tell me what is wrong with my code??

Answer 1

One problem I see is:

  char src[2] = {0, x};

You want to use the character 0, rather than the byte value 0 (NUL):

  char src[2] = {'0', x};

Also, you need to NUL terminate your src array:

  char src[] = {'0', x, 0};

The NUL termination would need to be done in both cases.

Another problem is your x = x - 96 statement also does not take into account that the character 0 is different from the byte value 0. So

x = x - 96 + '0';

would be better.

Answer 2

You use strcat() on an uninitialized array, result , it doesn't work that way, strcat() looks for the terminating nul byte in it's first parameter, and glues the second parameter from that point.

In your case there is no '\\0' in result, calling strcat() with result as the first parameter causes undefined behavior.

You should at least ensure that there is one '\\0' in result, you can do that by just setting the first element of result to '\\0' , like

result[0] = '\0';

Right before the loop starts, also you pass the second paramter src which is an array but is not a string, because it has no terminating '\\0' , but you actually don't need an array nor strcat() .

You can use an index variable and just assing the i th element of the array to the actual character computed from the original value, like

result[i] = x;

Then you don't nul terminate result , which causes undefined behavior when you try to print resutl .

You also passing a parameter to printf() where it's not expecting one, that indicates that you are either silencing or ignoring compiler warnings.

There are many problems in your code, the following code, does what you want

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

void s2n(char chr)
{
    char result[4];
    if (snprintf(result, sizeof(result), "%02d", 1 + chr - 'a') >= sizeof(result))
        return;
    printf("%s",  result);
}

int main(void)
{
    char chr;

    printf("Lower cased string please: ");
    if (scanf(" %c", &chr) != 1)
    {
        fprintf(stderr, "input error!\n");
        return -1;
    }
    s2n(chr);

    return 0;
}

note that no strings are required except the one where you are going to store the result, which you can build easily with sprintf() .