C：不兼容的类型？ [英] C: Incompatible types?

问题描述

#include <stdlib.h>
#include <stdio.h>

struct foo{
    int id;
    char *bar;
    char *baz[6];
};

int main(int argc, char **argv){
    struct foo f;   
    f.id=1;

    char *qux[6];

    f.bar=argv[0];
    f.baz=qux;  // Marked line

    return 1;
}

这只是一些测试code所以忽略qux实际上不会有任何在它有用。

This is just some test code so ignore that qux doesn't actually have anything useful in it.

我在标记线分配到类型的char * [6]从类型时，得到一个错误，不兼容的类型'字符** ，但两者的变量定义为的char * [6] 在code。任何见解？

I'm getting an error on the marked line, incompatible types when assigning to type ‘char *[6]’ from type ‘char **’ but both of the variables are defined as char *[6] in the code. Any insight?

推荐答案

本code的问题是该行

The problem with this code is that the line

f.baz=qux;

试图数组 qux 分配给数组巴兹。在C中，阵列是不变量，不能分配新值。要看到一个简单的例子，这个code：

Tries to assign the array qux to the array baz. In C, arrays are not variables and cannot be assigned new values. To see a simpler example, this code:

int arr1[10], arr2[10];
arr1 = arr2; // Error!

时非法的，因为在第二行试图第二阵列分配第一值，这是不是在C不允许

Is illegal because the second line tries to assign the second array the value of the first, which is not allowed in C.

要解决这个问题，你可能会想要写一个显式循环的元素拷贝过来，因为在这里看到：

To fix this, you'll probably want to write an explicit loop to copy the elements over, as seen here:

int i;
for (i = 0; i < 6; ++i)
    f.baz[i] = qux[i];

这是合法的，因为数组中的每个元素是一个的char * ，它可以分配。

This is legal because each individual element of the array is a char *, which can be assigned.

您有时还可看到的memcpy 用于指定数组：

You sometimes also see memcpy used to assign arrays:

memcpy(f.baz, qux, sizeof f.baz);

这比复制阵列中的所有字段的原始字节。

This copies over the raw bytes of all the fields in the arrays.

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