我知道在一些语言如下：
I know in some languages the following:
a += b的效率比：
a = a + b，因为它消除了对创建临时变量的需要。这是在C的情况下？它是更有效地使用+ =（因此也 - = * = 等）
推荐答案所以这里有一个明确的答案......
So here's a definitive answer...
$ cat junk1.c #include <stdio.h> int main() { long a, s = 0; for (a = 0; a < 1000000000; a++) { s = s + a * a; } printf("Final sum: %ld\n", s); } michael@isolde:~/junk$ cat junk2.c #include <stdio.h> int main() { long a, s = 0; for (a = 0; a < 1000000000; a++) { s += a * a; } printf("Final sum: %ld\n", s); } michael@isolde:~/junk$ for a in *.c ; do gcc -O3 -o ${a%.c} $a ; done michael@isolde:~/junk$ time ./junk1 Final sum: 3338615082255021824 real 0m2.188s user 0m2.120s sys 0m0.000s michael@isolde:~/junk$ time ./junk2 Final sum: 3338615082255021824 real 0m2.179s user 0m2.120s sys 0m0.000s...对于我计算机和我编译器上运行的我操作系统。结果可以或可以不变化。在我系统，但是，时间是相同的：用户时间2.120s
...for my computer and my compiler running on my operating system. Your results may or may not vary. On my system, however, the time is identical: user time 2.120s.
现在只是给你看pressive现代编译器如何IM就可以了，你会注意到，我使用了前pression 在A * A 分配。这是因为这个小问题：
Now just to show you how impressive modern compilers can be, you'll note that I used the expression a * a in the assignment. This is because of this little problem:
$ cat junk.c #include <stdio.h> int main() { long a, s = 0; for (a = 0; a < 1000000000; a++) { s = s + a; } printf("Final sum: %ld\n", s); } michael@isolde:~/junk$ gcc -O3 -S junk.c michael@isolde:~/junk$ cat junk.s .file "junk.c" .section .rodata.str1.1,"aMS",@progbits,1 .LC0: .string "Final sum: %ld\n" .text .p2align 4,,15 .globl main .type main, @function main: .LFB22: .cfi_startproc movabsq $499999999500000000, %rdx movl $.LC0, %esi movl $1, %edi xorl %eax, %eax jmp __printf_chk .cfi_endproc .LFE22: .size main, .-main .ident "GCC: (Ubuntu 4.4.3-4ubuntu5) 4.4.3" .section .note.GNU-stack,"",@progbits编译器想通了，我的循环，它展开来计算累积和的点，只是嵌入，作为一个常数，它进行打印出来，跳过任何一种循环完全建构的。在优化的脸聪明，你真的认为你会发现任何有意义的边缘 S =秒之间区别+ A 和 S + = A ？！
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为A + = B比在C = A + B更有效？
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