获取整数输入并在没有数组的情况下逐行输出

Question

当我使用 scanf 输入一个整数并拆分该整数并在没有数组的情况下逐行打印时，我正在尝试编写一个 c 程序。我可以举例说明我将如何做到这一点。

123 / 100 = 1
123 % 100 = 23

23 / 10 = 2
23 % 19 = 3

1
2
3

我知道怎么做，但问题是当我运行这个代码 .

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

int main (void)
{
    int no, a;
    int count, new;
    int newNum = 0;

    printf("Enter an intger number = ");
    scanf("%d", &no);

    newNum = no;
    printf("You entered = %d\n", newNum);

    while(newNum != 0){
        newNum = newNum / 10;
        count++;
    }

    count--;
    count = pow(10, count);

    printf("Power of ten = %d\n", count);

    while(count != 1){
        new = no / count;
        no = no % count;
        printf("%d\n", new);
        count = count / 10;
    }
    return 0;
}

输出：

Enter an intger number = 123
You entered = 123
Power of ten = -2147483648
0
0
0
0
0
0
0
0
-5
-9
Floating point exception (core dumped)

问题是十行的幂没有输出正确的值但是如果我评论第二个while循环部分。

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

int main (void)
{
    int no;
    int count, new;
    int newNum = 0;

    printf("Enter an intger number = ");
    scanf("%d", &no);

    newNum = no;
    printf("You entered = %d\n", newNum);

    while(newNum != 0){
        newNum = newNum / 10;
        count++;
    }

    count--;
    count = pow(10, count);

    printf("Power of ten = %d\n", count);

//  while(count != 1){
//      new = no / count;
//      no = no % count;
//      printf("%d\n", new);
//      count = count / 10;
//  }
    return 0;
}

输出：

Enter an intger number = 123
You entered = 123
Power of ten = 100

这次 10 的幂显示了正确的值。

我该怎么做才能避免这个问题？

和

有没有不用数组的方法？

Answer 1

在您的代码中，变量 count 具有局部作用域（自动存储持续时间）。

局部作用域变量由于其堆栈分配而未初始化。

int count=0;

（C99 标准）第 6.7.8 节第 10 条：

If an object that has automatic storage duration is not initialized explicitly, its value is indeterminate.

If an object that has static storage duration is not initialized explicitly, then:

if it has pointer type, it is initialized to a null pointer;

if it has arithmetic type, it is initialized to (positive or unsigned) zero;

if it is an aggregate, every member is initialized (recursively) according to these rules;

if it is a union, the ﬁrst named member is initialized (recursively) according to these rules.

注意：您应该避免使用 new 作为变量名。

Answer 2

对于初学者，您应该小心，不要在计算 10 的幂时溢出。

始终测试您的程序的边界值，例如 INT_MAX 或 UINT_MAX。

不需要使用数学函数。

给你。

#include <stdio.h>

int main( void ) 
{
    while ( 1 )
    {
        const unsigned int Base = 10;

        printf( "Enter a non-negative number (0 - exit): " );

        unsigned int n;

        if ( scanf( "%u", &n ) != 1 || n == 0 ) break;

        unsigned int divisor = 1; 

        for ( unsigned int tmp = n; !( tmp < Base ); tmp /= Base )
        {
            divisor *= Base;;
        }

        printf( "%u\n", n / divisor );

        for ( ; divisor != 1; divisor /= Base )
        {
            printf( "%u\n", n % divisor / ( divisor / Base ) );
        }

        putchar( '\n' );
    }

    return 0;
}

程序输出可能看起来像

Enter a non-negative number (0 - exit): 1
1

Enter a non-negative number (0 - exit): 10
1
0

Enter a non-negative number (0 - exit): 123456789
1
2
3
4
5
6
7
8
9

Enter a non-negative number (0 - exit): 4294967295
4
2
9
4
9
6
7
2
9
5

Enter a non-negative number (0 - exit): 0

另一种方法是使用此处所示的递归函数

#include <stdio.h>

void output_digits( unsigned int n )
{
    const unsigned int Base = 10;

    unsigned int digit = n % Base;

    if ( ( n /= Base ) != 0 ) output_digits( n );

    printf( "%u\n", digit );
}

int main( void ) 
{
    while ( 1 )
    {
        printf( "Enter a non-negative number (0 - exit): " );

        unsigned int n;

        if ( scanf( "%u", &n ) != 1 || n == 0 ) break;

        output_digits( n );

        putchar( '\n' );
    }

    return 0;
}

Answer 3

您想打印 10 进制数的每一位。可以用这个简单的公式

检索位数

int n_digit = (int)log10(x) +1;

测试：(int)log10(99) +1 = (int)1.9956 +1 = 2 和 (int)log10(100) +1 = (int)2 +1 = 3

所以你想打印 10 的每个幂的商：

for(int i=n_digit-1; i>=0; i--)
{
    // Calc 10^i without pow function
    int num_10_pow_i = 1;
    for(int j=0; j<i; j++)
        num_10_pow_i *=10;

    printf("N[%d]=",i);
    printf("%d\n", (int)(x/num_10_pow_i));
    x = x-((int)(x/num_10_pow_i) * num_10_pow_i);
}

请不要使用处理整数的 pow 函数。

阅读此处了解更多信息：

Answer 4

在我的程序中，我像这样将整数变量设置为 0。

int no, a = 0;
int count = 0, new = 0;
int newNum = 0;

之后我的程序运行完美，但没有得到完整的输出。这个问题来自这个部分

while(count != 1){
    new = no / count;
    no = no % count;
    printf("%d\n", new);
    count = count / 10;
}

输出-:

Enter an intger number = 123
You entered = 123
Power of ten = 100
1
2

你可以看到3 没有在输出部分。因为我在 while 循环之外添加了额外的行以获得我预期的输出。我已经添加了以下内容。

while(count != 1){
    new = no / count;
    no = no % count;
    printf("%d\n", new);
    count = count / 10;
}
printf("%d\n", no);

输出-:

Enter an intger number = 123
You entered = 123
Power of ten = 100
1
2
3

现在完成了。

获取整数输入并在没有数组的情况下逐行输出

Get Integer input and output it line by line without an array

c

loops

for-loop

while-loop

integer-division