C++中size_t含义疑问：malloc与array模板中的size_t是否等价？

Great question! Let's break this down clearly to eliminate any confusion:

1. The size_t type is identical in both scenarios #

First and foremost: yes, the size_t used in void *malloc(size_t _Size) and the std::array template template <class T, size_t N> class array is exactly the same standard type.

size_t is a core unsigned integer type defined in both C (via headers like <stddef.h>) and C++ (via <cstddef> or other standard library headers). It's explicitly designed to represent non-negative values related to size and count—like the byte size of an object, the number of elements in an array, or the length of a memory block. Its width matches the system's pointer size (32 bits on 32-bit systems, 64 bits on 64-bit systems), but its fundamental meaning is consistent across all standard-compliant C/C++ implementations.

2. The memory allocation difference comes from usage semantics, not size_t itself #

The reason your two examples allocate different amounts of memory has nothing to do with size_t itself—it's about what the size_t value represents in each context:

• For malloc(size_t _Size): The parameter _Size specifies the total number of bytes to allocate. So malloc(6*sizeof(int)) requests exactly enough bytes to store 6 int values (since sizeof(int) gives the byte size of a single int).
• For std::array<T, N>: The template parameter N specifies the number of elements the array will hold, not bytes. So array<int, 6*sizeof(int)> myArray creates an array that holds 6*sizeof(int) individual int elements. The total memory allocated is therefore N * sizeof(int), which equals 6*sizeof(int)*sizeof(int) bytes.

In short: size_t means the same thing everywhere, but its purpose (counting bytes vs counting elements) changes how we interpret its value in each use case.

内容的提问来源于stack exchange，提问作者xiezibuyongbi