Question

我正在查看以下程序，并且不确定如何分配内存以及原因：

void function() {
    char text1[] = "SomeText";
    char* text2 = "Some Text";
    char *text = (char*) malloc(strlen("Some Text") + 1 );
}

在上面的代码中，最后一个显然在堆中。但是，据我所知，text2位于程序的数据段中，而text1可能位于堆栈中。或者我的假设错了？这里的正确假设是什么？这个编译器是否依赖？

由于

Answer 1

// Array allocated on the stack and initialized with "SomeText" string.
// It has automatic storage duration. You shouldn't care about freeing memory.
char text1[] = "SomeText"; 

// Pointer to the constant string "Some Text".
// It has static storage duration. You shouldn't care about freeing memory.
// Note that it should be "a pointer to const".
// In this case you'll be protected from accidential changing of 
// the constant data (changing constant object leads to UB).
const char* text2 = "Some Text";

// malloc will allocate memory on the heap. 
// It has dynamic storage duration. 
// You should call "free" in the end to avoid memory leak.
char *text = (char*) malloc(strlen("Some Text") + 1 );

Answer 2

是的，你是对的，在大多数系统上：

text1将是堆栈上的可写变量数组（它必须是可写数组）

text2实际上必须是const char*，是的，它会指向可执行文件的文本段（但可能会在可执行格式中发生变化）

text将在堆

上

堆与数据段对堆栈分配

2 个答案: