pragma push/pop_macro指令的栈是否为每个宏独有？附示例求证

Great question! The short answer is: each macro has its own independent stack for push_macro/pop_macro operations—they don’t share a single global stack.

Let’s break down the provided example code to prove this point clearly:

First, here’s the code for reference:

// pragma_directives_pop_macro.cpp
// compile with: /W1
#include <stdio.h>
#define X 1
#define Y 2
int main() {
    printf("%d",X);
    printf(" %d",Y);
    #define Y 3 // C4005
    #pragma push_macro("Y")
    #pragma push_macro("X")
    printf(" %d",X);
    #define X 2 // C4005
    printf(" %d",X);
    #pragma pop_macro("X")
    printf(" %d",X);
    #pragma pop_macro("Y")
    printf(" %d",Y);
}

The output is: 1 2 1 2 1 3

Let’s walk through the key steps that confirm per-macro stacks:

• We start with X=1 and Y=2, which gives the first two output values.
• We redefine Y to 3, then push this new value of Y onto its dedicated stack.
• Next, we push the original value of X (1) onto its separate stack.
• We redefine X to 2 and print it—this only affects the current active value of X, not the stack associated with Y.
• When we pop X, it reverts to the pushed value (1), with zero impact on Y’s stored state.
• Finally, popping Y reverts it to the value we pushed (3), which doesn’t interfere with X’s current value at all.

If there was a single shared stack, we’d see cross-contamination between macro states, but the clean separation of X and Y throughout the execution confirms each macro manages its own stack of previous definitions.

内容的提问来源于stack exchange，提问作者Hrothgar