c++ - 堆与数据段与堆栈分配

Question

我正在查看以下程序，但不确定内存是如何分配的以及为什么：

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

在上面的代码中，最后一个显然是在堆中。但是，据我了解text2，它位于程序的数据段中，并且text1可能位于堆栈中。还是我的假设是错误的？这里正确的假设是什么？这个编译器依赖吗？

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（必须是可写数组）

text2实际上必须const char*是，是的，它将指向可执行文件的文本段（但这可能会因可执行格式而改变）

text将在堆上