警告:赋值从指针生成整数而不进行强制转换 [英] warning: assignment makes integer from pointer without a cast

问题描述

当我将 char * 声明为固定字符串并重用指针指向另一个字符串时

When I declare a char * to a fixed string and reuse the pointer to point to another string

/* initial declaration */
char *src = "abcdefghijklmnop";
.....

/* I get the   "warning: assignment makes integer from pointer without a cast" */
*src ="anotherstring";

我尝试重铸指针，但没有成功.

I tried to recast the pointer but no success.

推荐答案

写语句时

*src = "anotherstring";

编译器将常量字符串 "abcdefghijklmnop" 视为数组.假设您编写了以下代码:

the compiler sees the constant string "abcdefghijklmnop" like an array. Imagine you had written the following code instead:

char otherstring[14] = "anotherstring";
...
*src = otherstring;

现在，发生的事情更清楚了.左边的 *src 指的是一个 char(因为 src 是指向 char 的指针类型code>) 而右侧的 otherstring 指的是一个指针.

Now, it's a bit clearer what is going on. The left-hand side, *src, refers to a char (since src is of type pointer-to-char) whereas the right-hand side, otherstring, refers to a pointer.

这并不是严格禁止的，因为您可能想要存储指针指向的地址.但是，在这种情况下通常使用显式强制转换(这种情况不太常见).编译器发出危险信号，因为您的代码可能没有按照您的预期运行.

This isn't strictly forbidden because you may want to store the address that a pointer points to. However, an explicit cast is normally used in that case (which isn't too common of a case). The compiler is throwing up a red flag because your code is likely not doing what you think it is.

在我看来，您正在尝试分配一个字符串.C 中的字符串不是 C++ 中的数据类型，而是使用 char 数组实现的.您不能像尝试那样直接为字符串赋值.相反，您需要使用 strncpy 之类的函数和 <string.h> 中的朋友，并使用 char 数组而不是 char 指针.如果您只想让指针指向不同的静态字符串，请删除 *.

It appears to me that you are trying to assign a string. Strings in C aren't data types like they are in C++ and are instead implemented with char arrays. You can't directly assign values to a string like you are trying to do. Instead, you need to use functions like strncpy and friends from <string.h> and use char arrays instead of char pointers. If you merely want the pointer to point to a different static string, then drop the *.

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