寫這篇文章的原因是我在問答平台看到的一個問題:C++內層循環中定義變量和在外面定義比影響大嗎?
例如:
for(int i=0;i<999;i++) {
for(int j=0;j<999;j++);
}內層循環每次都定義j會造成多大的消耗呢?
此處我給出的回答是:
這個需要看你具體用什麼編譯器。不過主流編譯器(如vs和gcc)這一塊優化都比較好,不會反復分配變量。
看到答案和評論,好像有很多人對這個感興趣,所以我打算給大家實測分享一下,於是寫了如下代碼進行測試:
#include <cstdio>
using namespace std;
void Test1()
{
for (int i = 0; i < 2; i++)
{
for (int j = 0; j < 3; j++)
{
printf("%d,%d\n", int(i), int(j));
}
}
}
void Test2()
{
int i, j;
for (i = 0; i < 2; i++)
{
for (j = 0; j < 3; j++)
{
printf("%d,%d\n", int(i), int(j));
}
}
}
int main()
{
Test1();
Test2();
return 0;
}
OK,程序非常簡單,Test1
和Test2
是兩個循環,干相同的事情,就是在雙重循環裡打印一下 i
和 j
的值,差別只在於一個在循環外定義變量 j
,另一個在循環內定義變量 j
。
此處我使用g++
進行編譯,優化等級是O0
(這是GCC
默認的優化等級,也是最低的優化等級)的:
g++ -O0 -g test.cpp
編譯後,我將生成的Test1函數和Test2函數反匯編出來,得出的結果是這樣的:
Test1
函數反匯編如下:
(gdb) disas /m Test1
Dump of assembler code for function Test1():
5 {
0x0804841d <+0>: push %ebp
0x0804841e <+1>: mov %esp,%ebp
0x08048420 <+3>: sub $0x28,%esp
6 for (int i = 0; i < 2; i++)
0x08048423 <+6>: movl $0x0,-0x10(%ebp)
0x0804842a <+13>: jmp 0x804845d <Test1()+64>
0x08048459 <+60>: addl $0x1,-0x10(%ebp)
0x0804845d <+64>: cmpl $0x1,-0x10(%ebp)
0x08048461 <+68>: jle 0x804842c <Test1()+15>
7 {
8 for (int j = 0; j < 3; j++)
0x0804842c <+15>: movl $0x0,-0xc(%ebp)
0x08048433 <+22>: jmp 0x8048453 <Test1()+54>
0x0804844f <+50>: addl $0x1,-0xc(%ebp)
0x08048453 <+54>: cmpl $0x2,-0xc(%ebp)
0x08048457 <+58>: jle 0x8048435 <Test1()+24>
9 {
10 printf("%d,%d\n", int(i), int(j));
0x08048435 <+24>: mov -0xc(%ebp),%eax
0x08048438 <+27>: mov %eax,0x8(%esp)
0x0804843c <+31>: mov -0x10(%ebp),%eax
0x0804843f <+34>: mov %eax,0x4(%esp)
0x08048443 <+38>: movl $0x8048560,(%esp)
0x0804844a <+45>: call 0x80482f0 <printf@plt>
11 }
12 }
13
14 }
0x08048463 <+70>: leave
0x08048464 <+71>: ret
Test2
函數反匯編如下:
(gdb) disas /m Test2
Dump of assembler code for function Test2():
17 {
0x08048465 <+0>: push %ebp
0x08048466 <+1>: mov %esp,%ebp
0x08048468 <+3>: sub $0x28,%esp
18 int i, j;
19
20 for (i = 0; i < 2; i++)
0x0804846b <+6>: movl $0x0,-0x10(%ebp)
0x08048472 <+13>: jmp 0x80484a5 <Test2()+64>
0x080484a1 <+60>: addl $0x1,-0x10(%ebp)
0x080484a5 <+64>: cmpl $0x1,-0x10(%ebp)
0x080484a9 <+68>: jle 0x8048474 <Test2()+15>
21 {
22 for (j = 0; j < 3; j++)
0x08048474 <+15>: movl $0x0,-0xc(%ebp)
0x0804847b <+22>: jmp 0x804849b <Test2()+54>
0x08048497 <+50>: addl $0x1,-0xc(%ebp)
0x0804849b <+54>: cmpl $0x2,-0xc(%ebp)
0x0804849f <+58>: jle 0x804847d <Test2()+24>
23 {
24 printf("%d,%d\n", int(i), int(j));
0x0804847d <+24>: mov -0xc(%ebp),%eax
0x08048480 <+27>: mov %eax,0x8(%esp)
0x08048484 <+31>: mov -0x10(%ebp),%eax
0x08048487 <+34>: mov %eax,0x4(%esp)
0x0804848b <+38>: movl $0x8048560,(%esp)
0x08048492 <+45>: call 0x80482f0 <printf@plt>
25 }
26 }
27 }
0x080484ab <+70>: leave
0x080484ac <+71>: ret
End of assembler dump.
在Test1
的反匯編中,我們在內部for (int j = 0; j < 3; j++)
下面,沒有看到分配變量 j
的匯編指令,如果再只打印Test1
和Test2
的匯編代碼,經過對比,你們發現這兩個函數產生的匯編指令是完全一樣的:
(gdb) disas Test1
Dump of assembler code for function Test1():
0x0804841d <+0>: push %ebp
0x0804841e <+1>: mov %esp,%ebp
0x08048420 <+3>: sub $0x28,%esp
0x08048423 <+6>: movl $0x0,-0x10(%ebp)
0x0804842a <+13>: jmp 0x804845d <Test1()+64>
0x0804842c <+15>: movl $0x0,-0xc(%ebp)
0x08048433 <+22>: jmp 0x8048453 <Test1()+54>
0x08048435 <+24>: mov -0xc(%ebp),%eax
0x08048438 <+27>: mov %eax,0x8(%esp)
0x0804843c <+31>: mov -0x10(%ebp),%eax
0x0804843f <+34>: mov %eax,0x4(%esp)
0x08048443 <+38>: movl $0x8048560,(%esp)
0x0804844a <+45>: call 0x80482f0 <printf@plt>
0x0804844f <+50>: addl $0x1,-0xc(%ebp)
0x08048453 <+54>: cmpl $0x2,-0xc(%ebp)
0x08048457 <+58>: jle 0x8048435 <Test1()+24>
0x08048459 <+60>: addl $0x1,-0x10(%ebp)
0x0804845d <+64>: cmpl $0x1,-0x10(%ebp)
0x08048461 <+68>: jle 0x804842c <Test1()+15>
0x08048463 <+70>: leave
0x08048464 <+71>: ret
End of assembler dump.
(gdb) disas Test2
Dump of assembler code for function Test2():
0x08048465 <+0>: push %ebp
0x08048466 <+1>: mov %esp,%ebp
0x08048468 <+3>: sub $0x28,%esp
0x0804846b <+6>: movl $0x0,-0x10(%ebp)
0x08048472 <+13>: jmp 0x80484a5 <Test2()+64>
0x08048474 <+15>: movl $0x0,-0xc(%ebp)
0x0804847b <+22>: jmp 0x804849b <Test2()+54>
0x0804847d <+24>: mov -0xc(%ebp),%eax
0x08048480 <+27>: mov %eax,0x8(%esp)
0x08048484 <+31>: mov -0x10(%ebp),%eax
0x08048487 <+34>: mov %eax,0x4(%esp)
0x0804848b <+38>: movl $0x8048560,(%esp)
0x08048492 <+45>: call 0x80482f0 <printf@plt>
0x08048497 <+50>: addl $0x1,-0xc(%ebp)
0x0804849b <+54>: cmpl $0x2,-0xc(%ebp)
0x0804849f <+58>: jle 0x804847d <Test2()+24>
0x080484a1 <+60>: addl $0x1,-0x10(%ebp)
0x080484a5 <+64>: cmpl $0x1,-0x10(%ebp)
0x080484a9 <+68>: jle 0x8048474 <Test2()+15>
0x080484ab <+70>: leave
0x080484ac <+71>: ret
End of assembler dump.
當然,這裡只測試了g++
的編譯效果。vs
下的效果大家可以自己測試。目前可以肯定,如果你使用gcc
的編譯器,你完全可以不用糾結在循環外定義變量還是循環內定義變量,因為效果完全是一樣的,不過為了代碼好看,還是寫到循環內吧。
上面已經探究了使用基本數據類型int
作為循環變量的情況,這裡需要進階一下,探討一下如果我使用的不是int
,而是一個復雜的對象,那循環的效果又是如何呢?
為了方便看到變量的分配,我在類的構造函數裡加了打印語句,可以讓我們方便地看到類的對象被創建的情況:
#include <cstdio>
using namespace std;
class MyInt
{
public:
MyInt(int i):
m_iValue(i)
{
printf("Constructed: MyInt(%d)\n", i);
}
MyInt()
{
printf("Constructed: MyInt()\n");
}
MyInt &operator++(int i)
{
m_iValue ++;
return *this;
}
bool const operator <(const MyInt& another)
{
return m_iValue < another.m_iValue;
}
operator int()
{
return m_iValue;
}
MyInt &operator =(int i)
{
m_iValue = i;
return *this;
}
private:
int m_iValue;
};
void Test1()
{
for (MyInt i = MyInt(0); i < MyInt(2); i++)
{
for (MyInt j = MyInt(0); j < MyInt(3); j++)
{
printf("%d,%d\n", int(i), int(j));
}
}
}
void Test2()
{
MyInt i, j;
for (i = MyInt(0); i < MyInt(2); i++)
{
for (j = MyInt(0); j < MyInt(3); j++)
{
printf("%d,%d\n", int(i), int(j));
}
}
}
void Test3()
{
MyInt i, j;
for (i = 0; int(i) < 2; i++)
{
for (j = 0; int(j) < 3; j++)
{
printf("%d,%d\n", int(i), int(j));
}
}
}
int main()
{
printf("Test1---------------------------------\n");
Test1();
printf("Test2---------------------------------\n");
Test2();
printf("Test3---------------------------------\n");
Test3();
return 0;
}
好的,還是使用g++ -O0編譯,我們來看看執行結果:
Test1---------------------------------
Constructed: MyInt(0)
Constructed: MyInt(2)
Constructed: MyInt(0)
Constructed: MyInt(3)
0,0
Constructed: MyInt(3)
0,1
Constructed: MyInt(3)
0,2
Constructed: MyInt(3)
Constructed: MyInt(2)
Constructed: MyInt(0)
Constructed: MyInt(3)
1,0
Constructed: MyInt(3)
1,1
Constructed: MyInt(3)
1,2
Constructed: MyInt(3)
Constructed: MyInt(2)
Test2---------------------------------
Constructed: MyInt()
Constructed: MyInt()
Constructed: MyInt(0)
Constructed: MyInt(2)
Constructed: MyInt(0)
Constructed: MyInt(3)
0,0
Constructed: MyInt(3)
0,1
Constructed: MyInt(3)
0,2
Constructed: MyInt(3)
Constructed: MyInt(2)
Constructed: MyInt(0)
Constructed: MyInt(3)
1,0
Constructed: MyInt(3)
1,1
Constructed: MyInt(3)
1,2
Constructed: MyInt(3)
Constructed: MyInt(2)
Test3---------------------------------
Constructed: MyInt()
Constructed: MyInt()
0,0
0,1
0,2
1,0
1,1
1,2
可以看到,Test3
創建對象的次數是最少的,如果對象比較復雜,顯然Test3會是最高效的編碼方式。
對於整個程序的輸出,我們可以分析一下:
Test1
中仍然看到了對j變量多次分配動作。Test2
中,由於我們在循環外定義了j變量,所以這裡沒有發生對j變量的反復分配,但由於賦值條件i = MyInt(0)
和j = MyInt(0)
以及判斷條件i < MyInt(2)
和j < MyInt(3)
中需要構造MyInt(2)
和MyInt(3)
對象,所以我們仍然看到循環中多次的變量分配。Test3
中,我們換了一種方式,用重載運算符=
直接在賦值語句中給對象賦整型值,避免了賦值語句中創建MyInt
對象,並用int(i) < 2
和int(j) < 3
,避免了在判斷條件裡創建MyInt
對象,所以整段代碼裡只在循環外分配了兩次變量,這其實是最高效的方式。最後總結:
for
循環的賦值語句、判斷語句中,都要避免重復創建對象。------------------------------分割線------------------------------
C++ Primer Plus 第6版 中文版 清晰有書簽PDF+源代碼 http://www.linuxidc.com/Linux/2014-05/101227.htm
讀C++ Primer 之構造函數陷阱 http://www.linuxidc.com/Linux/2011-08/40176.htm
讀C++ Primer 之智能指針 http://www.linuxidc.com/Linux/2011-08/40177.htm
讀C++ Primer 之句柄類 http://www.linuxidc.com/Linux/2011-08/40175.htm
將C語言梳理一下,分布在以下10個章節中: