How to ignore/remove leading zeros?

Question

I am writing a program to add two large numbers in C. My integer array result holds the sum of the two numbers (which were also stored in arrays).

For example, if the result array is [0,0,3,2] (actual array size is 20)

If 32 is my actual result, how can I display the contents of the result array without the leading zeros ?

#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#define BASE 10
void align(int A[],int n);
void add(int A[],int B[], int C[]);
void Invert(int* a, int n);

int main(int argc, char** argv){
    char input1[20];
    char input2[20];
    int size = 20;
    int a;
    int b;
    int num1[20];
    int num2[20];
    int result[20];
    int length1 = strlen(argv[1]);
    int length2 = strlen(argv[2]);
    int i = 0;
    for (i=0;i<length1;i++){
       input1[i] = argv[1][i];
    }
    for (i=0;i<length2;i++){
        input2[i] = argv[2][i];
    }

    a=atoi(input1);
    b=atoi(input2);
    align(num1,a);
    align(num2,b);
    add(num1,num2,result);
    Invert(result,size);
    for (i=0;i<20;i++){
        printf("%d",result[i]);
    }

    return 0;
}

void align (int A[], int n){
    int i = 0;

    while (n) {
        A[i++] = n % BASE;

        n /= BASE;
    }

    while (i < 20) A[i++] = 0;
}

void add (int A[], int B[], int C[]) {
    int i, carry, sum;
    carry = 0;
    for (i=0; i<20; i++) {
        sum = A[i] + B[i] + carry;
        if (sum >= BASE) {
            carry = 1;
            sum -= BASE;
        } else
            carry = 0;
        C[i] = sum;
    }

    if (carry) printf ("overflow in addition!\n");
}

void Invert(int* a, int n)
{
    int i;
    int b;
    for(i=0; i<n/2; i++){
        b = a[i];
        a[i] = a[n-i-1];
        a[n-i-1] = b;
    } 
}

`

Solution 1

To get the actual digits (I assume that each digit is stored as a byte in an array of 20 bytes, lowest digit at highest index), you do something like this:

int i;
int size = sizeof(thearray) / sizeof(thearray[0]);

/* find first non-0 byte, starting at the highest "digit" */
for (i = 0; i < size - 1; ++i)
    if (thearray[i] != 0)
        break;

/* output every byte as character */
for (; i < size; i++)
    printf("%c", thearray[i] + '0'); /* 0 --> '0', 1 --> '1', etc. */
printf("\n");

Solution 2

You can do this by below code:-

int flag=1;
for(i=0;i<20;i++)
{
   if(flag==1&&array[i]!=0)
      flag=0;
   if(flag!=1)
   {
      printf("%d",array[i]);
   }
}

This will remove all leading zeros.

Comments

[simonc]

+1. Note that terminating the first loop at size-1 would allow you to avoid the later if else. Also, might be worth a note that you can only calculate size in the way you show if you have access to the array definition, not just a pointer to the array.

[Rudy Velthuis]

@simonc: you are right, the lowest digit can always be displayed. I'll change the code accordingly.

[simonc]

Won't this skip the first non-zero digit?

[A.s. Bhullar]

In case of 100 it will print 10

[A.s. Bhullar]

@simonc not a first non zero but first zero if zero is not leading, isn't it?

[Jongware]

Actually, for C and decimal digits the disclaimer "assuming ASCII" is obsolete. C mandates that the encoding of '0' to '9' digits is consecutive.

[Rudy Velthuis]

In C, chars and bytes can both be expressed as number or as character. Adding '0' (i.e. 0x30) to 0 turns the number 0 into the character '0' (ASCII 0x30), adding it to 1 returns the character '1' (ASCII 0x31), etc. The printf format "%c" expects a char, not a number. Alternatively, I could have used: printf("%d", thearray[i]);.

[Admin]

@RudyVelthuis Thanks !