SQRT（）函数不变量参数工作 [英] sqrt() function not working with variable arguments

问题描述

我不知道如果我失去了一些东西显而易见的，但现在看来，我无法计算的变量在C平方根;在SQRT（）函数只似乎​​对常数工作。这是我的code：

I don't know if I'm missing something obvious, but it appears that I'm unable to compute square roots of a variable in C; the sqrt() function only seems to work on constants. This is my code:

#include <math.h>
#include <stdio.h>

int main()
{
    double a = 2.0;
    double b = sqrt(a);
    printf("%f", b);
    return 0;
}

当我运行这个程序，我得到以下错误：

When I run this program, I get the following error:

gcc -Wall -o "test2" "test2.c" (in directory: /home/eddy/Code/euler)
/tmp/ccVfxkNh.o: In function `main':
test2.c:(.text+0x30): undefined reference to `sqrt'
collect2: ld returned 1 exit status
Compilation failed.

不过，如果我有一个常数这样的替代论点的sqrt（）为2.0，例如，（ B =开方（2.0）），那么它工作正常。是的sqrt（）不应该用变量或什么工作？

However, if I replace the argument in sqrt() with a constant such as 2.0 for example, (b = sqrt(2.0)), then it works fine. Is sqrt() not supposed to work with variables or something?

感谢您的帮助。

推荐答案

您需要（在命令行中使用'-lm'）与数学库链接。在不断的情况下，编译器可能是聪明和precomputing的sqrt（2.0）（以便在编译code是本质上是B = 1.414 ...;'）

You need to link with the math library (use a '-lm' on the command line). In the constant case, the compiler is probably being smart and precomputing sqrt(2.0) (so the code that is compiled is essentially 'b = 1.414...;')

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