Many times, the answer really depends on compiler - due to various optimizations.
So, best way is to compile the code, run it under gdb and inspect the assembly code. Here's what I got:
(gdb) b main
Starting program: /home/Mehul/a.exe
[New thread 4484.0x150c]
Breakpoint 1, 0x00401056 in main ()
Dump of assembler code for function main:
0x00401050 <main+0>: push %ebp
0x00401051 <main+1>: mov %esp,%ebp
0x00401053 <main+3>: sub $0x8,%esp
0x00401056 <main+6>: and $0xfffffff0,%esp
0x00401059 <main+9>: mov $0x0,%eax
0x0040105e <main+14>: add $0xf,%eax
0x00401061 <main+17>: add $0xf,%eax
0x00401064 <main+20>: shr $0x4,%eax
0x00401067 <main+23>: shl $0x4,%eax
0x0040106a <main+26>: mov %eax,-0x8(%ebp)
0x0040106d <main+29>: mov -0x8(%ebp),%eax
0x00401070 <main+32>: call 0x401084 <_alloca>
0x00401075 <main+37>: call 0x401114 <__main>
0x0040107a <main+42>: movl $0xa,-0x4(%ebp)
0x00401081 <main+49>: leave
0x00401082 <main+50>: ret
So, as you can see, after calling function main, 0xa is put on the stack. Memory is allocated on the stack.
But, let's tweak the source code:
int a, b;
b = a;
One would assume that stack would have two integers - however, if you compile with "gcc -o3 tmp.c".
After that, if you see the assembly code, it's exactly same as above for stack allocation.