警告：函数返回局部变量的地址[默认启用] [英] warning: function returns address of local variable [enabled by default]

问题描述

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

char *chktype(char *Buffer, int Size)
{
char *strng = "Content-Type: ";
int sz;
char *found = strstr (Buffer, strng);
char *found1 = strstr(found, "\r\n");
sz=strlen(found)-strlen(found1);
char type[sz];
strncpy(type, found1, sz-1);
return(type);
}
void main(){
char *buffer = "HTTP/1.1 200 OK\r\nDate: Tue, 25 Jun 2013 16:27:16
GMT\r\nExpires: -1\r\nCache-Control: private,
max-age=0\r\nContent-Type: text/html; 
charset=UTF-8\r\nContent-Encoding: gzip\r\nServer: 
gws\r\nX-XSS-Protection: 1; mode=block\r\nX-Frame-Options:
SAMEORIGIN\r\nTransfer-Encoding: chunked\r\n\r\n";
char *extension = chktype (buffer, sizeof(buffer));
printf("%s\r\n", extension);
}

//它给警告：函数返回局部变量的地址[由启用
默认]，我不能弄清楚什么是错误的，当我运行它，我希望可以将输出为的text / html;字符集= UTF-8 但其胡言乱语

推荐答案

chktype 函数分配内存，返回这个地址变量（即一个指向该变量）。

The chktype function allocates memory for an automatic variable on the stack, and then returns the address of this variable (i.e., a pointer to this variable).

问题是在栈上分配的变量时，他们走出去的范围时自动消亡（即，控制传递定义函数的花括号外面）。

The problem is that variables allocated on the stack are automatically destroyed whenever they go out of scope (i.e., control passes outside of the curly braces that define the function).

这意味着你基本上返回一个指针，指向一个无效的内存位置，这是个坏消息。在C-说话，这是的未定义行为的。在实际的讲，它会导致不好的输出或甚至崩溃。

This means that you're essentially returning a pointer to an invalid memory location, which is bad news. In C-speak, it's undefined behavior. In practical speak, it results in bad output or perhaps even a crash.

char *chktype(char *Buffer, int Size)
{
    // This pointer variable is allocated on the stack, but that's okay because
    // it's a pointer to a string literal, which are always constant.
    // (Technically, you should add the "const" qualifier to the declaration.)
    const char *strng = "Content-Type: ";

    int sz;
    char *found = strstr (Buffer, strng);
    char *found1 = strstr(found, "\r\n");
    sz=strlen(found)-strlen(found1);

    // Like all the above variables, the one is also allocated on the stack.
    // But it's the source of your problem here, because it's the one that
    // you are returning at the end of the function.
    // Problem is, it goes away at the end of the function!
    char type[sz];
    strncpy(type, found1, sz-1);
    return(type);
}

从函数返回一个的char * 正确的方法是使用的malloc （或释放calloc ）的功能。这意味着，在主叫的功能的将是负责释放由返回值所使用的内存，否则你的程序将导致内存泄漏。结果
的（始终把这一要求纳入你的函数的文档！即使文档是指上述声明的注释。）的

The correct way to return a char* from a function is to allocate new memory from the heap using the malloc (or calloc) function. That means that the caller of the function is going to be responsible for freeing the memory used by the returned value, otherwise your program will leak memory.
(Always put this requirement into the documentation for your function! Even if "documentation" means a comment above the declaration.)

例如，改变你的code看起来是这样的：

For example, change your code to look like this:

char *chktype(char *Buffer, int Size)
{
    // This pointer variable is allocated on the stack, but that's okay because
    // it's a pointer to a string literal, which are always constant.
    // (Technically, you should add the "const" qualifier to the declaration.)
    const char *strng = "Content-Type: ";

    int sz;
    char *found = strstr (Buffer, strng);
    char *found1 = strstr(found, "\r\n");
    sz=strlen(found)-strlen(found1);

    char *type = malloc(sz);  // allocate memory from the heap
    strncpy(type, found1, sz-1);
    return(type);
}

现在，在 chktype 函数的调用者，你必须确保你叫免费每当你其返回值完成：

Now, in the caller of the chktype function, you must make sure that you call free whenever you are finished with its return value:

char *type = chktype(...);
// do something
free(type);

需要注意的是稳健的code应该测试的malloc 为空指针，​​以确保它没有无法分配请求的内存的结果。如果是这样，你需要以某种方式处理错误。为清楚起见，未示出上面。

Note that robust code should test the result of malloc for a null pointer to make sure that it did not fail to allocate the requested memory. If so, you need to handle the error somehow. For clarity, that isn't shown above.

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