In C language, Why does n++
execute faster than n=n+1
?
(int n=...; n++;)
(int n=...; n=n+1;)
Our instructor asked
On GCC 4.4.3 for x86, with or without optimizations, they compile to the exact same assembly code, and thus take the same amount of time to execute. As you can see in the assembly, GCC simply converts n++
into n=n+1
, then optimizes it into the one-instruction add (in the -O2).
Your instructor's suggestion that n++
is faster only applies to very old, non-optimizing compilers, which were not smart enough to select the in-place update instructions for n = n + 1
. These compilers have been obsolete in the PC world for years, but may still be found for weird proprietary embedded platforms.
C code:
int n;
void nplusplus() {
n++;
}
void nplusone() {
n = n + 1;
}
Output assembly (no optimizations):
.file "test.c"
.comm n,4,4
.text
.globl nplusplus
.type nplusplus, @function
nplusplus:
pushl %ebp
movl %esp, %ebp
movl n, %eax
addl $1, %eax
movl %eax, n
popl %ebp
ret
.size nplusplus, .-nplusplus
.globl nplusone
.type nplusone, @function
nplusone:
pushl %ebp
movl %esp, %ebp
movl n, %eax
addl $1, %eax
movl %eax, n
popl %ebp
ret
.size nplusone, .-nplusone
.ident "GCC: (Ubuntu 4.4.3-4ubuntu5) 4.4.3"
.section .note.GNU-stack,"",@progbits
Output assembly (with -O2 optimizations):
.file "test.c"
.text
.p2align 4,,15
.globl nplusplus
.type nplusplus, @function
nplusplus:
pushl %ebp
movl %esp, %ebp
addl $1, n
popl %ebp
ret
.size nplusplus, .-nplusplus
.p2align 4,,15
.globl nplusone
.type nplusone, @function
nplusone:
pushl %ebp
movl %esp, %ebp
addl $1, n
popl %ebp
ret
.size nplusone, .-nplusone
.comm n,4,4
.ident "GCC: (Ubuntu 4.4.3-4ubuntu5) 4.4.3"
.section .note.GNU-stack,"",@progbits