How does the processor know the breakpoints?

Lets consider this very simple program:

#include<stdio.h>

int main () 
{
    int num1=4, num2=5;
    printf("Welcome\n");
    printf("num1 + num2 = %d\n", num1+num2);
    return 0;
}

When looking on the generated assembly code using gcc -S prog.c:

    .file   "p.c"
    .def    ___main;    .scl    2;  .type   32; .endef
    .section .rdata,"dr"
LC0:
    .ascii "Welcome\0"
LC1:
    .ascii "num1 + num2 = %d\12\0"
    .text
    .globl  _main
    .def    _main;  .scl    2;  .type   32; .endef
_main:
LFB10:
    .cfi_startproc
    pushl   %ebp
    .cfi_def_cfa_offset 8
    .cfi_offset 5, -8
    movl    %esp, %ebp
    .cfi_def_cfa_register 5
    andl    $-16, %esp
    subl    $32, %esp
    call    ___main
    movl    $4, 28(%esp)
    movl    $5, 24(%esp)
    movl    $LC0, (%esp)
    call    _puts
    movl    28(%esp), %edx
    movl    24(%esp), %eax
    addl    %edx, %eax
    movl    %eax, 4(%esp)
    movl    $LC1, (%esp)
    call    _printf
    call    _getchar
    movl    $0, %eax
    leave
    .cfi_restore 5
    .cfi_def_cfa 4, 4
    ret
    .cfi_endproc
LFE10:
    .ident  "GCC: (GNU) 5.3.0"
    .def    _puts;  .scl    2;  .type   32; .endef
    .def    _printf;    .scl    2;  .type   32; .endef
    .def    _getchar;   .scl    2;  .type   32; .endef

I know that the CPU see's the assembled code that the compiler generated for it, what I don't understand is how the program stops at the breakpoint that the user set ? why doesn't the CPU continue running the program ? what and how does that happens ? I mean, why would it stop after fetching an instruction ?

I am a bit confused about it, is it Code::Blocks caring for this or whatever program user is using ?

Thanks in advance!