Is x += 1 more efficient than x = x + 1?

Question

In x = x + 1, is x evaluated twice? If so, does that mean in x += 1, x is only evaluated once? How are the two expressions evaluated in terms of compiler intermediate code?

For example, x++ could mean : take the location of x, load the contents of x into a register, and increment the value of x in memory.

Also I have read that x += 1 is useful when x is not a simple variable, but an expression involving an array. Any ideas why this is the case?

Solution 1

In most compilers these would be identical

Solution 2

Just to give you a "real-world" example, consider this program:

int main()
{
    int i = 0;
    i += 1;
    i++;
    i = i + 1;
    return 0;
}

Compiling it with GCC, in Darwin 11 with the following flags:

• -S stop in assembler
• -m32 to 32-bit platform, just to simplify things a bit

Will generate the following program, except for the comments and blank lines which I added. Take a look specially in the comments.

        .section        __TEXT,__text,regular,pure_instructions
        .globl  _main
        .align  4, 0x90
_main:
        pushl   %ebp              # cdecl function stuff
        movl    %esp, %ebp        #
        subl    $12, %esp         # get room for variables    
        movl    $0, -12(%ebp)     # i = 0;

        ; i += 1
        movl    -12(%ebp), %eax   # load i in register a
        addl    $1, %eax          # add 1 to register a
        movl    %eax, -12(%ebp)   # store it back in memory

        ; i++
        movl    -12(%ebp), %eax   #
        addl    $1, %eax          # just the same
        movl    %eax, -12(%ebp)   #

        ; i = i + 1
        movl    -12(%ebp), %eax   #
        addl    $1, %eax          # just the same
        movl    %eax, -12(%ebp)   #

        movl    $0, -8(%ebp)
        movl    -8(%ebp), %eax
        movl    %eax, -4(%ebp)
        movl    -4(%ebp), %eax
        addl    $12, %esp
        popl    %ebp
        ret

.subsections_via_symbols

Solution 3

Why is x += 1 more efficient than x = x+1?

It isn't.

Why is x++ more efficient than x += 1?

It isn't.

---

The reason for preferring x += 1 to x = x+1 comes about when x is replaced with a much longer identifier, or perhaps a field in a class or struct. In that situation, the x += 1 version is more readable and even more importantly avoids the pitfalls of repeating yourself.

Solution 4

So there are already some questions that cover what you're asking here:

x=x+1 vs. x +=1

Incrementing: x++ vs x += 1

Which is faster? ++, += or x + 1?

The bottom line is that in most but not all languages the compiler is going to make them identical anyway, so there's no difference in efficiency. The questions above go into a fair bit of detail on the subject.

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