为什么指针不在堆栈上？

Question

void main(){
    int i,k;
    char* p;
    int j;
    printf("address of i is %d \naddress of k is %d \naddress of p is %p\naddress of j is %d", &i,&k,&p,&j);

}

when I tried the above code, the address of j is 4 units below k. But the address of p is no where near. Since a pointer is an integer variable that could store 4 bytes of data, why isn't it allocated on the stack like the other three variables?

Answer 1

Use %p to print all the addresses.

Spoiler: It is on the stack.

Answer 2

You should post the output you're getting. I suspect that you're getting a bit confused because most of the addresses you're printing are being displayed in decimal (using %d) while p is being displayed in hex (using %p).

Answer 3

I just tried your code on my computer (running Ubuntu 9.04) and got the following:

address of i is 0xbf96fe30
address of k is 0xbf96fe2c
address of p is 0xbf96fe28
address of j is 0xbf96fe24

after changing the code somewhat:

void main(){
    int i,k;
    char* p;
    int j;
    printf("address of i is %p \naddress of k is %p \naddress of p is %p\naddress of j is %p\n", &i,&k,&p,&j);

}

Since all you're printf() are addresses you should use %p instead of %d. Maybe you misinterpreted your results?

Answer 4

I can't help but tell you that your program is one of those that "should be indented six feet and covered with dirt." (google for who said it when and why:-) It is a prime example of sloppyness for the following reasons, most of which are causing undefined behavior:

• Uses printf without including stdio.h
• Uses void main
• Uses %d for addresses
• Uses %p with something other than a ptr-to-void

Properly written it looks like this:

#include <stdio.h>

int main (void)
{
    int i, k;
    char *p;
    int j;

    printf ("address of i is %p\n", (void *)&i);
    printf ("address of k is %p\n", (void *)&k);
    printf ("address of p is %p\n", (void *)&p);
    printf ("address of j is %p\n", (void *)&j);
    return 0;
}

Answer 5

When you use %d printf() specifier you get an address printed as decimal number and when you use %p specifier you get it printed as hexadecimal number. It just happened that the hexadecimal number contains no letters and you misinterpret it.

Answer 6

In the end a pointer is a variable.As far as the linux architecture is concerned,it is very clear that

All those variables whose memory is localized to a function will remain on the stack.<

There is no reason why for the above case the pointers should not be on the stack.Its a case of misinterpreting the ouput and ignoring the warning from gcc.Its a case with beginners that they ignore gcc warnings as harmless.There are cases when they are equivalent to a compiler error

Answer 7

Pointers may or may not be located on the stack as they are also some sort of variables. Note that they are some not very big variables. If the CPU/MCU has lots of registers and the compiler optimizes well, you may not see a pointer on the stack, it may very well spend its whole lifetime in registers.