I'm trying to print the number 684.545007 with 2 points precision in the sense that the number be truncated (not rounded) after 684.54.
我试图以6点精度打印数字684.545007,因为在684.54之后数字被截断(不是舍入)。
When I use
我用的时候
var = 684.545007;
printf("%.2f\n",var);
it outputs 684.55, but what I'd like to get is 684.54.
它输出684.55,但我想得到的是684.54。
Does anyone knows how can I correct this?
有谁知道我怎么能纠正这个?
3 个解决方案
#1
23
What you're looking for is truncation. This should work (at least for numbers that aren't terribly large):
您正在寻找的是截断。这应该工作(至少对于不是非常大的数字):
printf(".2f", ((int)(100 * var)) / 100.0);
The conversion to integer truncates the fractional part.
转换为整数会截断小数部分。
In C++11 or C99, you can use the dedicated function trunc for this purpose (from the header <cmath> or <math.h>. This will avoid the restriction to values that fit into an integral type.
在C ++ 11或C99中,您可以使用专用函数trunc(从头文件
std::trunc(100 * var) / 100 // no need for casts
#2
2
Here is my approach. It seems ugly but does work in most cases e.g. var can be larger then int, can be zero or bizarre '-0'. It does not handle infinities and NaNs though.
这是我的方法。它似乎很丑陋但在大多数情况下都有效,例如var可以大于int,可以是零或奇怪的'-0'。但它不处理无穷大和NaN。
double var = 684.545007; // or whatever
double var_trunc = var>=0. ? floor(var*100.)/100. : ceil(var*100.)/100.;
printf ("%g\n", var_trunc);
#3
1
printf("%.2f\n", var - 0.005);
#1
23
What you're looking for is truncation. This should work (at least for numbers that aren't terribly large):
您正在寻找的是截断。这应该工作(至少对于不是非常大的数字):
printf(".2f", ((int)(100 * var)) / 100.0);
The conversion to integer truncates the fractional part.
转换为整数会截断小数部分。
In C++11 or C99, you can use the dedicated function trunc for this purpose (from the header <cmath> or <math.h>. This will avoid the restriction to values that fit into an integral type.
在C ++ 11或C99中,您可以使用专用函数trunc(从头文件
std::trunc(100 * var) / 100 // no need for casts
#2
2
Here is my approach. It seems ugly but does work in most cases e.g. var can be larger then int, can be zero or bizarre '-0'. It does not handle infinities and NaNs though.
这是我的方法。它似乎很丑陋但在大多数情况下都有效,例如var可以大于int,可以是零或奇怪的'-0'。但它不处理无穷大和NaN。
double var = 684.545007; // or whatever
double var_trunc = var>=0. ? floor(var*100.)/100. : ceil(var*100.)/100.;
printf ("%g\n", var_trunc);
#3
1
printf("%.2f\n", var - 0.005);