不走出去的这样的范围内释放相关的内存？ [英] Does going out of scope like this free the associated memory?

问题描述

我只是想知道，在下面的scenarion，是方法1后释放'STRINGVAR'所使用的内存是做执行？

I was just wondering, in the following scenarion, is the memory used by 'stringvar' freed after method1 is done executing?

// Just some method
void method2(char* str)
{
  // Allocate 10 characters for str
  str = malloc(10 * sizeof(char));
}

// Just another method
void method1()
{
  char* stringvar;
  method2(stringvar);

  // Is the memory freed hereafter, or do I need to call free()?
}

我问，因为如果我把自由（STRINGVAR）的方法1月底，我得到一个警告，STRINGVAR在里面方法1（这是真的）未初始化。

I ask, because if I put a 'free(stringvar)' at the end of method1, I get a warning that stringvar is unitialized inside method1 (which is true).

推荐答案

没有，内存的不的方法1后释放，所以你'将有一个内存泄漏。是的，你需要调用免费您使用的内存中完成了。

No, the memory is not deallocated after method1, so you'll have a memory leak. Yes, you will need to call free after you're done using the memory.

您需要一个 指针发送到一个指针的以方法2 如果你想让它分配内存。这是C语言中常见的成语，尤其是当一个函数的返回值是保留整数状态codeS。例如，

You need to send a pointer to a pointer to method2 if you want it to allocate memory. This is a common idiom in C programming, especially when the return value of a function is reserved for integer status codes. For instance,

void method2(char **str) {
    *str = (char *)malloc(10);
}

char *stringvar;
method2(&stringvar);
free(stringvar);

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