C: how to break apart a multi digit number into se

问题:

Say I have a multi-digit integer in C. I want to break it up into single-digit integers.

123 would turn into 1, 2, and 3.

How can I do this, especially if I don't know how many digits the integer has?

回答1:

int value = 123;
while (value > 0) {
 int digit = value % 10;
 // do something with digit
 value /= 10;
}

回答2:

First, count the digits:

unsigned int count(unsigned int i) {
 unsigned int ret=1;
 while (i/=10) ret++;
 return ret;
}

Then, you can store them in an array:

unsigned int num=123; //for example
unsigned int dig=count(num);
char arr[dig];
while (dig--) {
 arr[dig]=num%10;
 num/=10;
}

回答3:

As a hint, getting the nth digit in the number is pretty easy; divide by 10 n times, then mod 10, or in C:

int nthdig(int n, int k){
     while(n--)
         k/=10;
     return k%10;
}

回答4:

The last digits of 123 is 123 % 10. You can drop the last digit of 123 by doing 123/10 -- using integer division this will give you 12. To answer your question about "how do I know how many digits you have" -- try doing it as described above and you will see how to know when to stop.

回答5:

I think below piece of code will help....

temp = num;
while(temp)
{
    temp=temp/10;
    factor = factor*10;
}

printf("\n%d\n", factor);
printf("Each digits of given number are:\n");

while(factor>1)
{
    factor = factor/10;
    printf("%d\t",num/factor);
    i++;
    num = num % factor;
}

回答6:

//Based on Tony's answer
#include <stdio.h> 
int nthdig(int n, int k){
    while(n--)
        k/=10;
    return k%10;
}

int main() {
    int numberToSplit = 987;
    printf("Hundreds = %i\n",nthdig(2, numberToSplit));
    printf("Tens     = %i\n",nthdig(1, numberToSplit));
    printf("Units    = %i\n",nthdig(0, numberToSplit));
}

This results in the following printout:

Hundreds = 9

Tens = 8

Units = 7

回答7:

I made this based on the code from @asaelr:

typedef struct digitsArrayPlusNumber {
    uint32_t *baseAddress;
    uint32_t number;
} digitsArrayPlusNumber;

digitsArrayPlusNumber *splitDigits (uint32_t inValue) {
    // based on code from [e-mail]

    uint32_t inputValue = inValue;

    //Count digits

    uint32_t theCount = 1;
    while (inputValue /= 10)
        theCount++;

    // put in array
    uint32_t *arr = malloc(sizeof(uint32_t) * theCount);
    uint32_t dig = theCount;
    while (dig--) {
        arr[dig]=inValue % 10;
        inValue /= 10;
        //  printf ("%d\n", arr[dig]);
    }

    digitsArrayPlusNumber *dandn = malloc (sizeof(digitsArrayPlusNumber));

    dandn->baseAddress = arr;
    dandn->number = theCount;

    return dandn;

}

int main(int argc, const char * argv[]) {


    for (int d = 0; d < splitDigits(12345678)->number; d++)
        printf ("%u\n", (splitDigits(12345678)->baseAddress)[d]);

}

It works quite well, thanks!

回答8:

You can use %10, which means the remainder if the number after you divided it. So 123 % 10 is 3, because the remainder is 3, substract the 3 from 123, then it is 120, then divide 120 with 10 which is 12. And do the same process.

回答9:

we can use this program as a function with 3 arguments.Here in "while(a++<2)", 2 is the number of digits you need(can give as one argument)replace 2 with no of digits you need. Here we can use "z/=pow(10,6)" if we don't need last certain digits ,replace 6 by the no of digits you don't need(can give as another argument),and the third argument is the number you need to break.

int main(){
long signed c=0,z,a=0,b=1,d=1;
scanf("%ld",&z);
while(a++<2){
       if(d++==1) 
       z/=pow(10,6);
       c+=(z%10)*b; 
       z/=10;
       b*=10;}
        return c;}

回答10:

You can divide and conquer but you have rewrite all of arithmetic libraries. I suggest using a multi-precision library https://gmplib.org But of course it is good practice

回答11:

int l1; //123456 for example
scanf("%d",&l1);
char s[sizeof(l1)];
sprintf(s,"%5d",l1);'

//This will give you separate digits of the number in char format inside s[0],s[1] 
//and so on.

//If you want them in int format, declare a int array say int i[sizeof(l1)] and add 
//the following code

for(int c=1;c<=sizeof(l1);c++){
i[c] = s[c] - '0';
} 

//Now i[0], i[1] etc will have the digits in int format