c - In ARM Cortex, the stack pointer increases instead of decreasing when allocating space for a local variable - Stack Overflow

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This is my C code

void user_main(void)
{
    u32 asd;

    osGetSysCnt(&asd);
}

This is the disassembly

void user_main(void)
{
10040230:   b507        push    {r0, r1, r2, lr}
    u32 asd;

    osGetSysCnt(&asd);
10040232:   a801        add r0, sp, #4
10040234:   f000 f801   bl  1004023a <osGetSysCnt>
}
10040238:   bd07        pop {r0, r1, r2, pc}

Why is the stack pointer incremented? Wouldn't that override the already pushed registers?

This is my C code

void user_main(void)
{
    u32 asd;

    osGetSysCnt(&asd);
}

This is the disassembly

void user_main(void)
{
10040230:   b507        push    {r0, r1, r2, lr}
    u32 asd;

    osGetSysCnt(&asd);
10040232:   a801        add r0, sp, #4
10040234:   f000 f801   bl  1004023a <osGetSysCnt>
}
10040238:   bd07        pop {r0, r1, r2, pc}

Why is the stack pointer incremented? Wouldn't that override the already pushed registers?

• c
• assembly
• arm
• stm32
• cortex-m

Share Improve this question Follow edited Nov 19, 2024 at 9:10 0___________ 68.5k44 gold badges3838 silver badges8686 bronze badges asked Nov 19, 2024 at 8:11 KorsarqKorsarq 79544 silver badges1919 bronze badges

Add a comment  | 

1 Answer 1

Sorted by: Reset to default 4

Why is the stack pointer incremented? Wouldn't that override the already pushed registers?

No, because sp is not incremented in the disassembly code.

10040232:   a801        add r0, sp, #4

Purpose of add: The add instruction calculates the memory address relative to the current value of sp (stack pointer). Here, add r0, sp, #4 computes the address sp + 4, which points to the location of the variable asd on the stack. The resulting address is stored in r0, which is then passed as a parameter to osGetSysCnt.

Isn't ARM stack full descending? Meaning sp+4 refers to the already pushed at the start registers.

It is. The push instruction is creating the stack frame sufficient for the function needs. The compiler does not care about what was there as ABI specifies that those registers do not have to be preserved. So this push is not preserving anything.

Why does the compiler use push instruction? It indicates that you are compiling with size optimizations enabled (-Os). push is short but time-consuming. If you change the optimization level to optimize for speed (-Ofast, -O2, -O3) the compiler will emit different code:

        str     lr, [sp, #-4]!
        ldr     r3, .L4
        sub     sp, sp, #12
        add     r0, sp, #4
        str     r3, [sp, #4]
        bl      osGetSysCnt
        add     sp, sp, #12
        ldr     lr, [sp], #4
        bx      lr

https://godbolt./z/7GY63r6vv

This is my C code

void user_main(void)
{
    u32 asd;

    osGetSysCnt(&asd);
}

This is the disassembly

void user_main(void)
{
10040230:   b507        push    {r0, r1, r2, lr}
    u32 asd;

    osGetSysCnt(&asd);
10040232:   a801        add r0, sp, #4
10040234:   f000 f801   bl  1004023a <osGetSysCnt>
}
10040238:   bd07        pop {r0, r1, r2, pc}

Why is the stack pointer incremented? Wouldn't that override the already pushed registers?

This is my C code

void user_main(void)
{
    u32 asd;

    osGetSysCnt(&asd);
}

This is the disassembly

void user_main(void)
{
10040230:   b507        push    {r0, r1, r2, lr}
    u32 asd;

    osGetSysCnt(&asd);
10040232:   a801        add r0, sp, #4
10040234:   f000 f801   bl  1004023a <osGetSysCnt>
}
10040238:   bd07        pop {r0, r1, r2, pc}

Why is the stack pointer incremented? Wouldn't that override the already pushed registers?

• c
• assembly
• arm
• stm32
• cortex-m

Share Improve this question Follow edited Nov 19, 2024 at 9:10 0___________ 68.5k44 gold badges3838 silver badges8686 bronze badges asked Nov 19, 2024 at 8:11 KorsarqKorsarq 79544 silver badges1919 bronze badges

Add a comment  | 

1 Answer 1

Sorted by: Reset to default 4

Why is the stack pointer incremented? Wouldn't that override the already pushed registers?

No, because sp is not incremented in the disassembly code.

10040232:   a801        add r0, sp, #4

Purpose of add: The add instruction calculates the memory address relative to the current value of sp (stack pointer). Here, add r0, sp, #4 computes the address sp + 4, which points to the location of the variable asd on the stack. The resulting address is stored in r0, which is then passed as a parameter to osGetSysCnt.

Isn't ARM stack full descending? Meaning sp+4 refers to the already pushed at the start registers.

It is. The push instruction is creating the stack frame sufficient for the function needs. The compiler does not care about what was there as ABI specifies that those registers do not have to be preserved. So this push is not preserving anything.

Why does the compiler use push instruction? It indicates that you are compiling with size optimizations enabled (-Os). push is short but time-consuming. If you change the optimization level to optimize for speed (-Ofast, -O2, -O3) the compiler will emit different code:

        str     lr, [sp, #-4]!
        ldr     r3, .L4
        sub     sp, sp, #12
        add     r0, sp, #4
        str     r3, [sp, #4]
        bl      osGetSysCnt
        add     sp, sp, #12
        ldr     lr, [sp], #4
        bx      lr

https://godbolt./z/7GY63r6vv

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