C语言 警告：传递参数“来自不兼容的指针类型[默认启用]”
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原文地址: http://stackoverflow.com/questions/14073870/
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warning: passing argument ’from incompatible pointer type [enabled by default]'
提问by mitit100
I've been looking around on the other threads somehow related to this, but somehow I just don't get it...
我一直在寻找与此相关的其他线程，但不知何故我只是不明白......
I want to do some FFT on a set of values I have evaluated and wrote this program to first read the values and save them to an array of size n.
我想对我已经评估过的一组值做一些 FFT 并编写这个程序来首先读取这些值并将它们保存到 size 的数组中n。
int main () {
// some variables and also a bit of code to read the 'messwerte.txt'
printf("Geben sie an wieviele Messwerte ausgelesen werden sollen: ");
scanf("%d", &n);
double werte[n]; //Array der "fertigen" Messwerte
in = fopen ("messwerte.txt","r");
double nul[n]; //Array von nullen
int logN = 14;
l=FFT(logN,&werte,&nul);
}
In the same file I also do the FFT with the help of this program:
在同一个文件中，我也在这个程序的帮助下做 FFT：
double FFT (int logN, double *real, double *im) //logN is base 2 log(N) {
// blabla FFT calculation
}
However, when I compile I always get this error:
但是，当我编译时，我总是收到此错误：
gcc FFT.c -lm
FFT.c: In function ‘main':
FFT.c:94:2: warning: passing argument 2 of ‘FFT' from incompatible pointer type [enabled by default]
FFT.c:4:8: note: expected ‘double *' but argument is of type ‘double (*)[(unsigned int)(n)]'
FFT.c:94:2: warning: passing argument 3 of ‘FFT' from incompatible pointer type [enabled by default]
FFT.c:4:8: note: expected ‘double *' but argument is of type ‘double (*)[(unsigned int)(n)]'
Since this is my first time programming, I really don't know what is wrong with my code. Will I have to set more flags for the compiler or stuff like that (because I had to do this -lmstuff or it wouldn't compile and said something like pow not found or so)?
由于这是我第一次编程，我真的不知道我的代码有什么问题。我是否必须为编译器或类似的东西设置更多标志（因为我必须做这些-lm东西，否则它不会编译并说诸如 pow not found 之类的东西）？
Also I was made aware that there might be a difference when writing on a Windows or a Linux machine, and I am using Linux, Lubuntu 12.10 32-bit if it's a problem of the OS.
我也意识到在 Windows 或 Linux 机器上编写代码可能会有所不同，如果是操作系统的问题，我使用的是 Linux、Lubuntu 12.10 32 位。
回答by cnicutar
l=FFT(logN,&werte,&nul); ^ ^
l=FFT(logN,&werte,&nul); ^ ^
Drop ampersands from that line.
从该行删除＆符号。
The problem is that the &operator in this context produces an expression with a different type than what FFTexpects. FFT expects a pointer to a double and &werteproduces a pointer to an array of N elements. So, in order to make FFThappy, just pass wertewhich will quietly decay to a pointer to the first element.
问题是&此上下文中的运算符生成的表达式与FFT预期的类型不同。FFT 需要一个指向 double 的指针，并&werte生成一个指向 N 个元素数组的指针。所以，为了让自己FFT开心，只需要传递werte一个会悄悄衰减到第一个元素的指针。
For more information on pointers to arrays, there's a C FAQ.
有关指向数组的指针的更多信息，请参阅C 常见问题解答。
回答by varevarao
werte[]and nul[]are arrays, but the word werteitself is an address of the first element of the array. So when you do &werteyou're trying to pass an address of the address (as @cnicutar pointed out, this should actually read pointer to an array of N elements). SO just pass werteand nulwithout the ampersand signs to pass the address of these arrays.
werte[]和nul[]是数组，但单词werte本身是数组第一个元素的地址。因此，当您这样做时，您&werte试图传递地址的地址（正如@cnicutar 指出的那样，这实际上应该读取指向 N 个元素数组的指针）。所以只需传递werte并且nul没有＆符号来传递这些数组的地址。
