C语言 C 复数和 printf
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C complex number and printf
提问by gameboy
How to print ( with printf ) complex number? For example, if I have this code:
如何打印（使用 printf ）复数？例如，如果我有这个代码：
#include <stdio.h>
#include <complex.h>
int main(void)
{
double complex dc1 = 3 + 2*I;
double complex dc2 = 4 + 5*I;
double complex result;
result = dc1 + dc2;
printf(" ??? \n", result);
return 0;
}
..what conversion specifiers ( or something else ) should I use instead "???"
..我应该使用什么转换说明符（或其他）来代替“???”
回答by John Calsbeek
printf("%f + i%f\n", creal(result), cimag(result));
I don't believe there's a specific format specifier for the C99 complex type.
我不相信 C99 复杂类型有特定的格式说明符。
回答by levif
Let %+fchoose the correct sign for you for imaginary part:
让我们%+f为您的虚部选择正确的符号：
printf("%f%+fi\n", crealf(I), cimagf(I));
Output:
输出：
0.000000+1.000000i
Note that iis at the end.
注意i是在最后。
回答by Kipp Cannon
Because the complex number is stored as two real numbers back-to-back in memory, doing
因为复数作为两个实数在内存中背靠背存储，所以
printf("%g + i%g\n", result);
will work as well, but generates compiler warnings with gcc because the type and number of parameters doesn't match the format. I do this in a pinch when debugging but don't do it in production code.
也可以工作，但会使用 gcc 生成编译器警告，因为参数的类型和数量与格式不匹配。调试时我会在紧要关头这样做，但不要在生产代码中这样做。
回答by Lou
Using GNU C, this works:
使用 GNU C，这有效：
printf("%f %f\n", complexnum);
Or, if you want a suffix of "i" printed after the imaginary part:
或者，如果您想在虚部后打印“i”后缀：
printf("%f %fi\n", complexnum);
