为什么我的 char 数组没有在 C 中打印出来？

Question

I have this program that is trying to convert a character string to a different base without using atoi. Why isn't the output printing at the end? The text "Output:" prints, but there is nothing after it.

The code seems to work up until that point. I went ahead and added the null byte.

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

void Int(char* numstring, int base) {

    char *digitPtr = numstring;
    char ans[100];

    int decVal = 0;
    int totalVal = 0;

    int numDigits = 0;

    // Go through the entire string to get the decimal value of the number.

    while (*digitPtr != '\0') {
        numDigits++;
        digitPtr++;
    }

    // Now calculate decimal value of the string.

    printf("before decimal value calc \n");
    //  *digitPtr = numstring;

    int length = numDigits;

    numDigits--;

    printf("numDigits: %d \n", numDigits);

    
    for (int i = 0; i < length; i++)
    {
        printf("calcuating the dec \n");
        decVal = (*numstring - 48);
        printf("dec Value: %d \n", decVal);
        printf("power: %f \n", pow(10, numDigits));     

        totalVal = totalVal + decVal *  pow(10, numDigits);

        printf("totalVal %d \n", totalVal);
        numDigits--;
        numstring++; 
    }

    //*digitPtr = numstring;
    numDigits = length;

    printf("totalVal %d \n", totalVal);

    while (totalVal > 0) {
        int remainder = totalVal % base;
        printf("remainder: %d \n", remainder);
        totalVal = totalVal / base;
        ans[numDigits] = remainder;
        numDigits--;
    }

    ans[99] = '\0'; 

    printf("Output: %s \n", ans); 
}

int main() {

Answer 1

It doesn't print anything because you're assigning the numerical value of the remainder to a char , but you're not considering that numbers start at 48 (that is, '0' ) in the ASCII table, so you should add the "starting point" to your remainder:

ans[numDigits] = remainder + '0';

That said, there's another problem: you're assigning the length of the input value to numDigits before the while loop, but the length of the number won't probably be the same as length of the number in the new base (for example, 32 in base 10 is two digits long, but in base 2 it has four more digits). A solution could be to make numDigits start at 0, increment it after every iteration and print the string character by character in reverse order. Alternatively a mathematical approach for taking the number of digits of a number in a custom base is adding one to the rounded-down logarithm of the number in the custom base. To do so you can you the properties of logarithm since C standard libraries only provide logs in base 2, e, and 10.

numDigits = (int)(floor(log(totalVal) / log(base) + 1);

Following this approach you don't have to change your loop and you can print the result as a string.

Note that you shouldn't set ans[99] to '\0'; since there could be garbage between the converted number and the end of the string; instead, you should set it to numDigits . You could also initialize ans to 0 (which is the numerical representation of \0 ), so you can forget about manually setting the terminator.