在C ++中不输出尾随零 [英] Do Not Output Trailing Zeroes in C++

问题描述

如果结果是整数，则不输出小数点.

If the result is an integer, do not output the decimal point.

如果结果是浮点数，则不要输出任何尾随零.

If the result is a floating point number, do not output any trailing zeroes.

如何在 c 或 c ++ 中进行操作:

double result;
/*some operations on result */
printf("%g", result);

以上方法正确吗?在我的情况下，我没有得到正确的答案.

Is the above approach correct? I'm not getting the correct answer with in my situation.

推荐答案

您可以使用 snprintf 将double值打印到char数组中，然后从头到头，替换为'0'加上'\ 0'，最后得到一个没有尾部零的数字.这是我的简单代码.

you can use snprintf to print the double value into a char array, then from the end to the head, replace the '0' with '\0', finally you get a the number without tailling zeroes.Here is my simple code.

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

void
remove_zeroes(double number, char * result, int buf_len)
{ 
    char * pos;
    int len;

    snprintf(result, buf_len, "%lf", number);
    len = strlen(result);

    pos = result + len - 1;
 #if 0
/* according to Jon Cage suggestion, removing this part */
    while(*div != '.')
        div++;
#endif
    while(*pos == '0')
        *pos-- = '\0';

    if(*pos == '.')
        *pos = '\0';

}

int
main(void)
{
    double a;
    char test[81];

    a = 10.1;

    printf("before it is %lf\n", a);
    remove_zeroes(a, test, 81);
    printf("after it is %s\n", test);

    a = 100;
    printf("before it is %lf\n", a);
    remove_zeroes(a, test, 81);
    printf("after it is %s\n", test);


    return 0;
}

输出为

before it is 10.100000
after it is 10.1
before it is 100.000000
after it is 100

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