C语言 如何在C中将浮点数与整数相乘？
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How to multiply float with integers in C?
提问by Wicelo
When I execute this code it returns me 1610612736
当我执行此代码时，它返回给我 1610612736
void main(){
float a=3.3f;
int b=2;
printf("%d",a*b);
}
Why and how to fix this ?
为什么以及如何解决这个问题？
edit : It's not even a matter of integer and float, if i replace int b=2: by float b=2.0f it return the same silly result
编辑：这甚至不是整数和浮点数的问题，如果我替换 int b=2: by float b=2.0f 它返回相同的愚蠢结果
回答by Carl Norum
The result of the multiplication of a floatand an intis a float. Besides that, it will get promoted to doublewhen passing to printf. You need a %a, %e, %for %gformat. The %dformat is used to print inttypes.
afloat和 an相乘的结果int是 a float。除此之外，它会double在传递给printf. 你需要一个%a，%e，%f或%g格式。该%d格式用于打印int类型。
Editorial note: The return value of mainshould be int. Here's a fixed program:
编者按：的返回值main应该是int。这是一个固定的程序：
#include <stdio.h>
int main(void)
{
float a = 3.3f;
int b = 2;
printf("%a\n", a * b);
printf("%e\n", a * b);
printf("%f\n", a * b);
printf("%g\n", a * b);
return 0;
}
and its output:
及其输出：
$ ./example
0x1.a66666p+2
6.600000e+00
6.600000
6.6
回答by krthkr
Alternately, you could also do
或者，你也可以这样做
printf("%d\n", (int)(a*b));
and this would print the result you're (kind of) expecting.
这将打印您（有点）期望的结果。
You should always explicitly typecast the variables to match the format string, otherwise you could see some weird values printed.
您应该始终明确地对变量进行类型转换以匹配格式字符串，否则您可能会看到一些奇怪的值被打印出来。
