C/C++中cout和printf的运行机制

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在进行分析之前先看一个例子：

例子1：

#include <iostream>#include <stdio.h>using namespace std;static int x=3;int foo1(){   x+=2;   cout<<"the first value of x is:"<<x<<endl;   return x;}int foo2(){   x+=2;   cout<<"the second value of x is:"<<x<<endl;   return x;} int main(){   int i = 0;   cout<<"cout the value: "<<"i="<<i<<' '<<"i++="<<i++<<' '<<"i--="<<i--<<endl;   printf("printf the value: i= %d, i++= %d, i--= %d\n",i,i++,i--);   cout<<"the case 1: "<<endl;   cout<<"the value of foo1: "<<foo1()<<endl;   cout<<"the value of foo2: "<<foo2()<<endl;      return 0;}

例子2：

#include <iostream>#include <stdio.h>using namespace std;static int x=3;int foo1(){   x+=2;   cout<<"the first value of x is:"<<x<<endl;   return x;}int foo2(){   x+=2;   cout<<"the second value of x is:"<<x<<endl;   return x;} int main(){   int i = 0;   cout<<"cout the value: "<<"i="<<i<<' '<<"i++="<<i++<<' '<<"i--="<<i--<<endl;   printf("printf the value: i= %d, i++= %d, i--= %d\n",i,i++,i--);      cout<<"the case 2:"<<endl;   cout<<"\t the value of foo1: "<<foo1()<<"\t"<<"the value of foo2: "<<foo2()<<endl;   return 0;}

观察例子1和例子2，唯一的不同就是cout输出foo1和foo2的值；这两个例子体现了cout和printf输出的运行机制，首先先看下这两个例子的输出：

//例子1输出：/*cout the value: i=0 i++=-1 i--=0printf the value: i= 0, i++= -1, i--= 0the case 1:the first value of x is:5the value of foo1: 5the second value of x is:7the value of foo2: 7*///例子2输出：/*cout the value: i=0 i++=-1 i--=0printf the value: i= 0, i++= -1, i--= 0the case 2:the second value of x is: 5the first value of x is: 7the value of foo1: 7the value of foo2: 5*//*从两个例子的输出可以看出foo1和foo2输出不同，下面将对cout和printf的输出机制进行分析，进行解释printf和cout的输出问题。*/

在c/c++中，cout和printf的机制是先从右往左读入缓冲区，然后再从左往右输出；

#include <iostream>using namespace std;int main(){   int a = 7;   int b = 6;   int c =5;     cout<<"a= "<<a<<' '<<"b= "<<b<<' '<<"c= "<<c<<endl;   return 0;}//cout和printf是先从右往左读入缓冲区，然后再从左往右输出；/*buffer:| 5 | 6 | 7 | 即依次把c=5,b=6,a=7压入缓冲区output:| 7 | 6 | 5 | 即依次输出a=7,b=6,c=5;*/

有了对cout和printf输出机制的了解，我们就可以分析上面例子的输出的结果：

//注：x是static变量，初始值x=3； int i = 0;   cout<<"cout the value: "<<"i="<<i<<' '<<"i++="<<i++<<' '<<"i--="<<i--<<endl;/*上面输出cout the value: i=0 i++=-1 i--=0；因为从右到左依次压入缓冲区，最右边的i=0压入缓冲区，再进行i--，则i=-1；继续把i=-1压入缓冲区，再进行i++，此时i=0；最后把i=0压入缓冲区；则此时缓冲区的值为buffer:| 0 | -1 | 0 |;输出时从左到右依次是0，-1,0；*/   printf("printf the value: i= %d, i++= %d, i--= %d\n",i,i++,i--);/*上面输出printf the value: i=0 i++=-1 i--=0；因为从右到左依次压入缓冲区，最右边的i=0压入缓冲区，再进行i--，则i=-1；继续把i=-1压入缓冲区，再进行i++，此时i=0；最后把i=0压入缓冲区；则此时缓冲区的值为buffer:| 0 | -1 | 0 |;输出时从左到右依次输出i,i++,i--是0，-1,0；因此cout和printf的运行机制是一样的*/   cout<<"the case 1: "<<endl;   cout<<"the value of foo1: "<<foo1()<<endl;/*上面语句cout<<"the value of foo1: "<<foo1()<<endl;首先调用foo1()函数，在foo1()函数中，先进行x+=2，即x=5；然后执行cout<<"the first value of x is: "<<endl;即在屏幕打印the first value of x is:5；继而执行return x；返回值为5，所以cout<<"the value of foo1: "<<foo1()输出the value of foo1: 5;*/   cout<<"the value of foo2: "<<foo2()<<endl;/*上面语句cout<<"the value of foo2: "<<foo2()<<endl;首先调用foo2()函数，在foo2()函数中，先进行x+=2，即x=7；然后执行cout<<"the second value of x is: "<<endl;即在屏幕打印the second value of x is:7；继而执行return x；返回值为7，所以cout<<"the value of foo2: "<<foo2()输出the value of foo2: 7;*///下面分析case 2cout<<"the case 2:"<<endl;   cout<<"\t the value of foo1: "<<foo1()<<"\t"<<"the value of foo2: "<<foo2()<<endl;/*    上面语句首先调用foo2()函数，再调用foo1()函数；在foo2()函数中，先进行x+=2，即x=5；然后执行cout<<"the second value of x is: "<<endl;即在屏幕打印the second value of x is:5；继而执行return x；返回值为5，cout时把foo2()函数的返回值5压入缓冲区；    进而调用foo1()函数，在foo1()函数中，先进行x+=2，即x=7；然后执行cout<<"the first value of x is: "<<endl;即在屏幕打印the first value of x is:7；继而执行return x；返回值为7，cout时把foo1()函数的返回值7压入缓冲区；此时缓冲区的值buffer：| 5 | 7 |；则从左到右依次输出foo1,foo2的值为7,5*/

最后给出一个例子：

#include <iostream>using namespace std;int foo(){   static int x =3;   x+=2;   cout<<"the value of x is:"<<x<<endl;   return x;} int main(){   cout<<"the value of foo: "<<foo()<<"\t"<<foo()<<endl;   return 0;}/*output:the value of x is:5the value of x is:7the value of foo: 75*/

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