是否可以找到分配给指针的内存，而无需搜索 malloc 语句 [英] Is it possible to find the Memory Allocated to the Pointer, without searching for the malloc statement

问题描述

假设我已经为函数 foo 中的某个指针分配了内存:

Suppose I have allocated memory to some pointer in a function foo:

void foo()
{    
    // ...  
    int *ptr = malloc(20*sizeof(int));  

    bar (ptr);
}  

从 foo()，我将这个指针传递给 bar()，让我们说从 bar() 到另一个函数.

From foo(), I pass this pointer to bar() and let's say from bar() to another function.

现在，在某个时间点，我想检查:为指针分配了多少内存.

Now, at some point of time, I want to check: How much memory was allocated to the pointer.

有没有可能的方法，不用搜索语句:

Is there any possible way, without searching for the statement:

int *ptr = malloc(20*sizeof(int)); 

使用 GDB 计算分配给指针的内存量?

to figure out how much memory is allocated to the pointer, using GDB?

谢谢.

推荐答案

答案是:它依赖.

许多系统提供 msize() [1]、malloc_usable_size() [2] 或类似功能.如果你在这样的系统上，(gdb) print malloc_usable_size(ptr) 就是你所需要的.

Many systems provide msize() [1], malloc_usable_size() [2], or similar function. If you are on such a system, (gdb) print malloc_usable_size(ptr) is all you need.

[1] http://msdn.microsoft.com/en-us/library/z2s077bc(v=vs.80).aspx
[2] http://www.slac.stanford.edu/comp/unix/package/rtems/doc/html/libc/libc.info.malloc.html

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