为什么在单个语句中调用两个函数不会影响值？[重复]

在此代码中：

// Stack using LinkedList //

#include <stdio.h>
#include <stdlib.h>

struct Node {
    int data;
    struct Node* next;
};

struct Node* top = NULL;

short isEmpty(void) {
    if (top == NULL) {
        printf("error: The stack is empty.\n");
        return 1;
    }
    
    return 0;
}

void push(int value) {
    struct Node* newNode = (void*)malloc(sizeof(struct Node));
    if (!newNode) {
        printf("error: Heap overflow!\n");
        exit(1);
    }
    
    newNode->data = value;
    newNode->next = top;
    top = newNode;
}

int pop(void) {
    if (isEmpty()) {
        exit(1);
    }
    
    struct Node* ref = top;
    int val = top->data;
    top = top->next;
    free(ref);
    
    return val;
}

int peek(void) {
    if (isEmpty()) {
        exit(1);
    }
    
    return top->data;
}

void display(void) {
    if (isEmpty()) {
        exit(1);
    }
    
    while (top) {
        printf("%d\n", top->data);
        top = top->next;
    }
}

int main(void) {
    push(10);
    push(20);
    push(30);
    push(40);
    
    printf("Peek: %d\n", peek());
    int val = pop();
    printf("Popped: %d, now Peek: %d\n", val, peek());
    push(50);
    display();
    
    return 0;
}

看看这些行：

int val = pop();
printf("Popped: %d, now Peek: %d\n", val, peek());

返回：Popped: 40, now Peek: 30

正如预期的那样。但是，当这样写时：

printf("Popped: %d, now Peek: %d\n", pop(), peek());

它产生以下输出：Popped: 40, now Peek: 40

这是Godbolt。

有人能告诉我为什么会这样吗？