从一个整数中得到每个数字

使用模运算符:

while(score)
{
    printf("%d\n", score % 10);
    score /= 10;
}

请注意,这将以相反的顺序给出数字(即先给出最低有效位)。如果你想先得到最重要的数字,你必须将这些数字存储在一个数组中,然后以相反的顺序读出它们。

同意前面的回答。

一点修正:有一种更好的方法可以从左到右打印十进制数字,而不需要分配额外的缓冲区。此外,如果 score

int div;
for (div = 1; div <= score; div *= 10)
    ;

do
{
    div /= 10;
    printf("%d\n", score / div);
    score %= div;
} while (score);

不要重新发明轮子。C有 sprintf

因为变量名为score,所以我猜这是一个游戏,您计划使用分数的各个数字将数字图示符显示为图像。在这种情况下, 把格式数据写成串

此解决方案在整个范围内给出正确的结果[0,UINT\u MAX] 无需缓冲数字。

它还适用于更广泛的类型或有符号类型(具有正值),并进行适当的类型更改。

#include <limits.h>
#include <stdio.h>

int
main (void)
{
  unsigned int score = 42;   // Works for score in [0, UINT_MAX]

  printf ("score via printf:     %u\n", score);   // For validation

  printf ("score digit by digit: ");
  unsigned int div = 1;
  unsigned int digit_count = 1;
  while ( div <= score / 10 ) {
    digit_count++;
    div *= 10;
  }
  while ( digit_count > 0 ) {
    printf ("%d", score / div);
    score %= div;
    div /= 10;
    digit_count--;
  }
  printf ("\n");

  return 0;
}

通常,这个问题可以通过在循环中使用数字的模来解决,或者将数字转换为字符串。要将数字转换为字符串,可以使用函数 itoa ,所以考虑循环中一个数的模的变量。

文件的内容 get_digits.c

$ cat get_digits.c 

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


// return a length of integer
unsigned long int get_number_count_digits(long int number);

// get digits from an integer number into an array
int number_get_digits(long int number, int **digits, unsigned int *len);

// for demo features
void demo_number_get_digits(long int number);


int
main()
{
    demo_number_get_digits(-9999999999999);
    demo_number_get_digits(-10000000000);
    demo_number_get_digits(-1000);
    demo_number_get_digits(-9);
    demo_number_get_digits(0);
    demo_number_get_digits(9);
    demo_number_get_digits(1000);
    demo_number_get_digits(10000000000);
    demo_number_get_digits(9999999999999);
    return EXIT_SUCCESS;
}


unsigned long int
get_number_count_digits(long int number)
{
    if (number < 0)
        number = llabs(number);
    else if (number == 0)
        return 1;

    if (number < 999999999999997)
        return floor(log10(number)) + 1;

    unsigned long int count = 0;
    while (number > 0) {
        ++count;
        number /= 10;
    }
    return count;
}


int
number_get_digits(long int number, int **digits, unsigned int *len)
{
    number = labs(number);

    // termination count digits and size of a array as well as
    *len = get_number_count_digits(number);

    *digits = realloc(*digits, *len * sizeof(int));

    // fill up the array
    unsigned int index = 0;
    while (number > 0) {
        (*digits)[index] = (int)(number % 10);
        number /= 10;
        ++index;
    }

    // reverse the array
    unsigned long int i = 0, half_len = (*len / 2);
    int swap;
    while (i < half_len) {
        swap = (*digits)[i];
        (*digits)[i] = (*digits)[*len - i - 1];
        (*digits)[*len - i - 1] = swap;
         ++i;
    }

    return 0;
}


void
demo_number_get_digits(long int number)
{
    int *digits;
    unsigned int len;

    digits = malloc(sizeof(int));

    number_get_digits(number, &digits, &len);

    printf("%ld --> [", number);
    for (unsigned int i = 0; i < len; ++i) {
        if (i == len - 1)
            printf("%d", digits[i]);
        else
            printf("%d, ", digits[i]);
    }
    printf("]\n");

    free(digits);
}

用GNU GCC演示

$~/Downloads/temp$ cc -Wall -Wextra -std=c11 -o run get_digits.c -lm
$~/Downloads/temp$ ./run
-9999999999999 --> [9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9]
-10000000000 --> [1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0]
-1000 --> [1, 0, 0, 0]
-9 --> [9]
0 --> [0]
9 --> [9]
1000 --> [1, 0, 0, 0]
10000000000 --> [1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0]
9999999999999 --> [9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9]

用LLVM/Clang演示

$~/Downloads/temp$ rm run
$~/Downloads/temp$ clang -std=c11 -Wall -Wextra get_digits.c -o run -lm
setivolkylany$~/Downloads/temp$ ./run
-9999999999999 --> [9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9]
-10000000000 --> [1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0]
-1000 --> [1, 0, 0, 0]
-9 --> [9]
0 --> [0]
9 --> [9]
1000 --> [1, 0, 0, 0]
10000000000 --> [1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0]
9999999999999 --> [9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9]

$~/Downloads/temp$ cc --version | head -n 1
cc (Debian 4.9.2-10) 4.9.2
$~/Downloads/temp$ clang --version
Debian clang version 3.5.0-10 (tags/RELEASE_350/final) (based on LLVM 3.5.0)
Target: x86_64-pc-linux-gnu
Thread model: posix

//this can be easily understandable for beginners     
int score=12344534;
int div;
for (div = 1; div <= score; div *= 10)
{

}
/*for (div = 1; div <= score; div *= 10); for loop with semicolon or empty body is same*/
while(score>0)
{
    div /= 10;
    printf("%d\n`enter code here`", score / div);
    score %= div;
}

#include<stdio.h>

int main() {
int num; //given integer
int reminder;
int rev=0; //To reverse the given integer
int count=1;

printf("Enter the integer:");
scanf("%i",&num);

/*First while loop will reverse the number*/
while(num!=0)
{
    reminder=num%10;
    rev=rev*10+reminder;
    num/=10;
}
/*Second while loop will give the number from left to right*/
while(rev!=0)
{
    reminder=rev%10;
    printf("The %d digit is %d\n",count, reminder);
    rev/=10;
    count++; //to give the number from left to right 
}
return (EXIT_SUCCESS);}

sprintf chars unsigned 分数:

unsigned int score = 1529587, i;
char stringScore [11] = { 0 };

sprintf( stringScore, "%d, score );

for( i=0; i<strlen(stringScore); i++ )
    printf( "%c\n", stringScore[i] );

请注意:

• stringScore 假设 int ,在您的平台中为4字节,因此最大整数长度为10位。第十一个是字符串终止符 '\0' .
• 把格式数据写成串 为你做所有的工作

你需要每一个数字都有一个整数吗?

stringScore公司 只包含数字,转换非常简单。如果 dig

int intDigit = dig - '0';

我已经做了这个解决方案,它很简单,取而代之的是读取一个整数,我读取一个字符串(C中的char数组),然后用一个for bucle来写,代码还写数字的和

// #include<string.h>

scanf("%s", n);
int total = 0;

for (int i = 0; i< strlen(n); i++){
    printf("%c", n[i]);
    total += (int)(n[i]) -48;
}

printf("%d", total);