了解指针和局部范围内 | 了解指针和局部范围内

本文介绍了了解指针和局部范围内的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！

问题描述

假设我有以下功能：

char* allocateMemory()
{
    char str[20] = "Hello world.";
    return str;
}

int* another()
{
    int x = 5;
    return &x;
}

int _tmain(int argc, _TCHAR* argv[])
{
    char* pString = allocateMemory();
    printf("%s\n", pString);

    int* blah = another();
    printf("%d %d \n", blah, *blah);

    return 0;
}

第一个printf打印随机值，因为str是本地作用域。

The first printf prints random values, because str IS LOCAL SCOPE.

第二个printf打印出正确的价值观，有嗒嗒=嗒嗒的地址，*等等= 5

The second printf prints the proper values, with blah = address of blah, *blah = 5

为什么本地范围仅影响allocateMemory它与数组的交易，而不是整数？

Why is it that local scope only affects allocateMemory which deals with arrays, but not integer?

为什么第一个printf（返回的char *）打印随机值，由局部范围内的影响，但不是第二个（返回int *）？

Why does the first printf (returning char* ) prints random values and is affected by local scope, but not the second one (returning int* )?

推荐答案

访问这超出范围是未定义行为的方法的局部变量的两种方式。这些都是一些有效的方式：

Both ways of accessing the local variables of a method which goes out of scope is Undefined Behavior. These are some valid ways:

char* allocateMemory()
{
    char* str= malloc(sizeof(char) * 20); //assuming C
    strcpy(str, "Hello World.");
    return str; //Valid
}

const char* allocateMemory()
{
    return "Hello world."; //Valid Hello World is in read only location
}

int* another()
{
    int *x = malloc(sizeof(int)); //assuming C
    *x = 5;
    return x; //Valid
}

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