如何格式化c程序中的％e格式说明符 [英] How to format the %e format specifier in c program

问题描述

float num = 10.0111

printf（％E，num）;



输出：1.001110E + 001



但我想要= 1.00111E + 1



我尝试了什么：



我读了许多网站，有人写过C库中没有提供解决方案。

float num = 10.0111
printf("%E",num);

Output: 1.001110E+001

But I want = 1.00111E+1

What I have tried:

I read many website there were written that no hands on solution is available in C library.

推荐答案

行为由标准定义（参见 fprintf [ ^ ]）：

The behaviour is defined by the standard (see fprintf[^]):

Quote：

指数应始终包含至少两位数。如果该值为零，则指数应为零。

The exponent shall always contain at least two digits. If the value is zero, the exponent shall be zero.

因此，唯一的解决方案是使用 sprintf（）打印到缓冲区，然后修改字符串。



示例：

So the only solution would be printing to a buffer using sprintf() and modifying the string afterwards.

Example:

char buffer[64];
sprintf(buffer, "%E", num);
char *expPos = strchr(buffer, 'E');
int exp = atoi(expPos + 1);
sprintf(expPos + 1, "%+d", exp);
printf(buffer);

删除尾随零的版本在指数之前：

A version that removes also trailing zeroes before the exponent:

char buffer[64];
sprintf(buffer, "%E", num);
// Locate the exponent
char *expPos = strchr(buffer, 'E');
// Get the exponent value
int exp = atoi(expPos + 1);
// Skip trailing zeroes using pre-increment to start at the last digit.
// Afterwards expPos points to the last non-zero digit
do expPos--; while (expPos > buffer && *expPos == '0');
// Overwrite trailing zeroes and exponent with short exponent
sprintf(expPos + 1, "E%+d", exp);
printf(buffer);

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