C语言 打印没有小数位的双精度数
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Printing a double without the decimal places
提问by phoenix
I have a number stored as a doubleand want to print it without the decimal places. So if I had the double value 919545634521.000000, it is always printed with the decimal places added to it. How can I print it without it so it looks like: 919545634521?
我有一个存储为 a 的数字double，想打印它而不带小数位。所以如果我有 double value 919545634521.000000，它总是在打印时加上小数位。如果没有它，我如何打印它，使其看起来像：919545634521？
#include<stdlib.h>
int main()
{
double number = 9220343120;
printf("%??\n", number);
}
回答by Michael Krelin - hacker
try
尝试
printf("%.0lf\n",phoneNum);
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long long phoneNum;
phoneNum = strtoll(buffer,NULL,0);
printf("%lld\n",phoneNum);
instead. Depending on the system, though, you may need other function to convert (I think it's _strtoui64for windows).
反而。但是，根据系统的不同，您可能需要其他函数来转换（我认为它_strtoui64适用于 Windows）。
回答by NPE
Well, don't store the phone number as a floating-point number (one wrong move, and you'll end up with your telephone numbers getting roundedfor you).
好吧，不要将电话号码存储为浮点数（一个错误的举动，您最终会为您的电话号码四舍五入）。
Store it as an integer or a string.
将其存储为整数或字符串。
回答by Richard Holland
回答by jman
printf ("%.0f\n", phoneNum);should work.
printf ("%.0f\n", phoneNum);应该管用。
回答by caf
You're looking for precision- the format specifier %.0fshould do what you want.
您正在寻找精度- 格式说明符%.0f应该做你想做的。
回答by nevelis
Not sure about other compiler support, but GCC supports the 'g' format specifier, which will print a double using the least space required for full precision. For example:
不确定其他编译器是否支持，但 GCC 支持 'g' 格式说明符，它将使用全精度所需的最少空间打印双精度值。例如：
double d1 = 10.000;
double d2 = 25.03;
printf( "d1: %g\nd2: %g\n", d1, d2 );
will result in:
将导致：
d1: 10
d2: 25.03
