C-打印浮点值 [英] C - Printing out float values

问题描述

我有一个C ++程序，它接受值并打印出这样的值:

I have a C++ program that takes in values and prints out values like this:

getline(in,number);
cout << setw(10) << number << endl;

我有一个等效的C程序，它接受值并像这样打印:

I have an equivalent C program that takes in values and prints out like so:

fscanf(rhs, "%e", &number);
printf("%lf\n", number);

但是在C ++程序打印出 0.30951 的同时，C程序打印出 0.309510 .更多示例:C ++: 0.0956439 C: 0.095644 .只要该值是7位数字，它看起来就可以打印出相同的结果，但是如果它的7位数字较短，它将在末尾添加一个额外的0.如果长度超过7位，则四舍五入为6位.我希望C结果与C ++程序匹配.任何帮助将不胜感激.

But while the C++ program prints out, 0.30951 the C program prints out 0.309510. More examples: C++: 0.0956439 C: 0.095644. It seems to print the same results as long as the value is 7 digits long, but if its shorter the 7 digits, it adds an extra 0 at the end. And if its longer than 7 digits, it rounds down to 6 digits. I would like the C results to match the C++ program. Any help would be appreciated.

谢谢.

注意:数字是浮点数，数字是从文件中读取的.

Note: number is a float and number are read from a file.

推荐答案

在C格式的打印语句中利用长度和精度说明符:

Take advantage of the length and precision specifiers in C formatted print statements:

printf( "%6.4lf", number );

在一个宽度为六个字符的单元格"中打印四个小数位.

Prints four decimal places in a "cell" six characters wide.

您可以使用通配符来指定长度或精度，以在运行时提供该值:

You can use a wildcard character for either length or precision to provide that value at runtime:

int precision = 4;

printf( "%6.*lf", precision, number );

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