使用 printf("%s",&Buffer[i]); 在 C

Question

void Store(int temp){
    Buffer[i]=temp;
    i--;

}

int main(int argc, char *argv[]){
    Buffer[64]=00;

    int value = atoi(argv[1]);
    int base = atoi(argv[2]);

    char Table[16] = { '0', '1', '2', '3', '4', '5' , '6', '7', '8', '9', 'A', 'B', 'C', 'D', 'E', 'F'};

    //Mathmatical algorithm 
    while(value !=0){   
        digit = value%base;
        value = value/base;
        Store(digit);
        printf("%s",&Buffer[i]);
    }

I'm starting to learn C on my own. I'm trying to figure out how to use printf with this char Buffer[65]. Essentially, my algorithm pulls in a value and base given by the user, then calculates what that number is in that base. If I change my printf to print out digit, it prints "123" for 57 base 4. So, I have it storing digit into the Buffer array in reverse (set int i to 63 because i set 64 to null).

Long story short, when it stores into the buffer, increments down, then returns to the while loop it doesn't print any text.

Edit: As I'm looking around, it looks like printf only prints ascii characters. So I need to convert digit to one of the characters in Table[16]?

Answer 1

Here's a working answer, though it only works up to base 16:

#include <stdio.h>
char Buffer[65];
int i = 63;

void Store(int temp){
    Buffer[i]=temp;
    i--;    
}

int main(int argc, char *argv[]){
    Buffer[64]=00;

    int value = atoi(argv[1]);
    int base = atoi(argv[2]);

    printf( "%d in base %d = ", value, base );

    char Table[16] = { '0', '1', '2', '3', '4', '5' , '6', '7', '8', '9', 'A', 'B', 'C', 'D', 'E', 'F'};

    //Mathmatical algorithm
    while(value !=0){
        int digit = value%base;
        value = value/base;
        Store( Table[digit]);  // or: Buffer[i--] = Table[digit];
    }
    printf("%s\n", &Buffer[i+1]);
}