答案很简单……一个非常了解汇编的人(也就是他身边有参考资料，并利用每一个小处理器缓存和管道特性等)保证能够产生比任何编译器更快的代码。

然而，如今在典型的应用程序中，这种差异并不重要。

这很难具体地回答，因为这个问题非常不具体:到底什么是“现代编译器”?

理论上，几乎任何手动的汇编器优化都可以由编译器来完成——实际上它是否已经完成，不能笼统地说，只能说特定编译器的特定版本。许多可能需要花费大量的精力来确定它们是否可以在特定的上下文中应用而不产生副作用，以至于编译器编写者不会为它们烦恼。

尽管C语言“接近”于对8位、16位、32位和64位数据的低级操作，但仍有一些C语言不支持的数学操作通常可以在某些汇编指令集中优雅地执行:

Fixed-point multiplication: The product of two 16-bit numbers is a 32-bit number. But the rules in C says that the product of two 16-bit numbers is a 16-bit number, and the product of two 32-bit numbers is a 32-bit number -- the bottom half in both cases. If you want the top half of a 16x16 multiply or a 32x32 multiply, you have to play games with the compiler. The general method is to cast to a larger-than-necessary bit width, multiply, shift down, and cast back: int16_t x, y; // int16_t is a typedef for "short" // set x and y to something int16_t prod = (int16_t)(((int32_t)x*y)>>16);` In this case the compiler may be smart enough to know that you're really just trying to get the top half of a 16x16 multiply and do the right thing with the machine's native 16x16multiply. Or it may be stupid and require a library call to do the 32x32 multiply that's way overkill because you only need 16 bits of the product -- but the C standard doesn't give you any way to express yourself. Certain bitshifting operations (rotation/carries): // 256-bit array shifted right in its entirety: uint8_t x[32]; for (int i = 32; --i > 0; ) { x[i] = (x[i] >> 1) | (x[i-1] << 7); } x[0] >>= 1; This is not too inelegant in C, but again, unless the compiler is smart enough to realize what you are doing, it's going to do a lot of "unnecessary" work. Many assembly instruction sets allow you to rotate or shift left/right with the result in the carry register, so you could accomplish the above in 34 instructions: load a pointer to the beginning of the array, clear the carry, and perform 32 8-bit right-shifts, using auto-increment on the pointer. For another example, there are linear feedback shift registers (LFSR) that are elegantly performed in assembly: Take a chunk of N bits (8, 16, 32, 64, 128, etc), shift the whole thing right by 1 (see above algorithm), then if the resulting carry is 1 then you XOR in a bit pattern that represents the polynomial.

尽管如此，除非有严重的性能限制，否则我不会求助于这些技术。正如其他人所说，汇编代码比C代码更难记录/调试/测试/维护:性能的提高伴随着一些严重的代价。

编辑:3。溢出检测在汇编中是可能的(在C中不能真正做到)，这使得一些算法更容易。

我认为汇编程序更快的一般情况是，当一个聪明的汇编程序员看到编译器的输出并说“这是性能的关键路径，我可以写这个更有效”，然后那个人调整汇编程序或从头重写它。

几乎任何时候编译器看到浮点代码，如果你使用的是旧的糟糕的编译器，手写的版本会更快。(2019年更新:对于现代编译器来说，这并不普遍。特别是在编译x87以外的东西时;编译器更容易使用SSE2或AVX进行标量数学运算，或任何具有平面FP寄存器集的非x86，不像x87的寄存器堆栈。)

主要原因是编译器不能执行任何健壮的优化。关于这个主题的讨论，请参阅来自MSDN的这篇文章。下面是一个例子，其中汇编版本的速度是C版本的两倍(用VS2K5编译):

#include "stdafx.h"
#include <windows.h>

float KahanSum(const float *data, int n)
{
   float sum = 0.0f, C = 0.0f, Y, T;

   for (int i = 0 ; i < n ; ++i) {
      Y = *data++ - C;
      T = sum + Y;
      C = T - sum - Y;
      sum = T;
   }

   return sum;
}

float AsmSum(const float *data, int n)
{
  float result = 0.0f;

  _asm
  {
    mov esi,data
    mov ecx,n
    fldz
    fldz
l1:
    fsubr [esi]
    add esi,4
    fld st(0)
    fadd st(0),st(2)
    fld st(0)
    fsub st(0),st(3)
    fsub st(0),st(2)
    fstp st(2)
    fstp st(2)
    loop l1
    fstp result
    fstp result
  }

  return result;
}

int main (int, char **)
{
  int count = 1000000;

  float *source = new float [count];

  for (int i = 0 ; i < count ; ++i) {
    source [i] = static_cast <float> (rand ()) / static_cast <float> (RAND_MAX);
  }

  LARGE_INTEGER start, mid, end;

  float sum1 = 0.0f, sum2 = 0.0f;

  QueryPerformanceCounter (&start);

  sum1 = KahanSum (source, count);

  QueryPerformanceCounter (&mid);

  sum2 = AsmSum (source, count);

  QueryPerformanceCounter (&end);

  cout << "  C code: " << sum1 << " in " << (mid.QuadPart - start.QuadPart) << endl;
  cout << "asm code: " << sum2 << " in " << (end.QuadPart - mid.QuadPart) << endl;

  return 0;
}

和一些数字从我的PC运行默认版本*:

  C code: 500137 in 103884668
asm code: 500137 in 52129147

出于兴趣，我用dec/jnz交换了循环，它对计时没有影响——有时更快，有时更慢。我想内存有限的方面使其他优化相形见绌。(编者注:更可能的情况是，FP延迟瓶颈足以隐藏循环的额外成本。对奇数/偶数元素并行进行两个Kahan求和，并在最后添加它们，可能会加快2倍的速度。)

哎呀，我正在运行一个稍微不同的代码版本，它输出的数字是错误的(即C更快!)修正并更新了结果。

一个更著名的组装片段来自Michael Abrash的纹理映射循环(在这里详细解释):

add edx,[DeltaVFrac] ; add in dVFrac
sbb ebp,ebp ; store carry
mov [edi],al ; write pixel n
mov al,[esi] ; fetch pixel n+1
add ecx,ebx ; add in dUFrac
adc esi,[4*ebp + UVStepVCarry]; add in steps

现在，大多数编译器将高级CPU特定指令表示为intrinsic，即编译为实际指令的函数。MS Visual c++支持MMX、SSE、SSE2、SSE3和SSE4的intrinsic，因此您不必太过担心使用特定于平台的指令来进行汇编。Visual c++还可以通过适当的/ARCH设置来利用您所针对的实际体系结构。