C - 打印出浮点值
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C - Printing out float values
提问by FidelCashflo
I have a C++ program that takes in values and prints out values like this:
我有一个 C++ 程序，它接受值并打印出这样的值：
getline(in,number);
cout << setw(10) << number << endl;
I have an equivalent C program that takes in values and prints out like so:
我有一个等效的 C 程序，它接受值并像这样打印出来：
fscanf(rhs, "%e", &number);
printf("%lf\n", number);
But while the C++ program prints out, 0.30951the C program prints out 0.309510. More examples: C++: 0.0956439C: 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.
但是当 C++ 程序打印出来时，0.30951C 程序打印出0.309510. 更多示例：C++: 0.0956439C: 0.095644. 只要值是 7 位长，它似乎打印相同的结果，但如果它的 7 位数字更短，它会在末尾添加一个额外的 0。如果长度超过 7 位，则向下舍入为 6 位。我希望 C 结果与 C++ 程序匹配。任何帮助，将不胜感激。
Thanks.
谢谢。
Note: number is a float and number are read from a file.
注意：数字是一个浮点数，数字是从文件中读取的。
回答by Bob Kaufman
Take advantage of the length and precision specifiers in C formatted print statements:
利用 C 格式打印语句中的长度和精度说明符：
printf( "%6.4lf", number );
Prints four decimal places in a "cell" six characters wide.
在宽度为六个字符的“单元格”中打印四位小数。
You can use a wildcard character for either length or precision to provide that value at runtime:
您可以使用通配符作为长度或精度来在运行时提供该值：
int precision = 4;
printf( "%6.*lf", precision, number );
回答by Alex Chamberlain
Take advantage of the length and precision specifiers in C++ iostreams
利用长度和精度说明的在C ++iostream小号
std::cout.precision(4);
std::cout << std::setw(10) << number << "\n";
回答by Sava Vrane?evi?
printf("%g\n", number);
Will solve your problem, %lf is used for double, %f is used for float, and %g is used for float when you want to display all the decimal places and cut off zeros.
将解决您的问题， %lf 用于 double ， %f 用于浮点数， %g 用于浮点数，当您要显示所有小数位并截断零时。
