Stroustrup 5.9，练习4（第105页） [英] Stroustrup 5.9, exercise 4 (page 105)
问题描述
这个程序编译但遇到了一些奇怪的结果，一个语义 -
BUG就在那里。我甚至放了一些std :: cout要知道出了什么问题，但它确实是b $ b做了一些我没有命令它做的事情：
------------ ---计划-------------------
/ * Stroustrup，5.9运动4
声明：
编写一个交换2个整数的程序。解决它
使用1st指针然后引用。
方法：
1.）要求用户输入2个整数
2.）使用指针交换价值，
打印2个语句以显示它正在做什么
3.）打印SWAPPED值
4.）使用参考资料交换谷物评价
5.）所以在交换2次之后我们得到原始输入值。
* /
#include< iostream>
void swap_p（int * x，int * y）;
void swap_r（int& x，int& y ）;
int main（）
{
使用std :: cout;
使用std :: cin;
使用std :: endl;
int i;
int j;
cout<< 输入第一个整数：;
cin> i;
cout<< 输入第二个整数：;
cin> j;
cout<< i = << i<< ''\t''
<< j = << j<< endl;
int * pi =& i;
int * pj =& j;
swap_p（pi， pj）;
cout<< " \ nvalues使用Poiners \ n交换
<< i = << i<< ''\t''
<< j = << j<< endl;
swap_r（i，j）;
cout<< " \ nswappe dusing Refrences\\\
"
<< i = << i<< ''\t''
<< j = << j<< endl;
返回0;
}
void swap_p（int * x，int * y）
{
int temp_val = * x;
x = y;
std :: cout<< " ------------------------------ \ n";
std :: cout< ;< x = << * x<< " \ty =" << * y<<''\ n'';
* y = temp_val;
std :: cout<< x = << * x<< " \ty =" << * y<<''\ n'';
std :: cout<< ------------------------------ \ n;
}
void swap_r（int& x，int& y）
{
int temp_val = x;
x = y;
y = temp_val;
}
--------------输出----- -------------
[arch @ voodo tc ++ pl] $ g ++ -ansi -pedantic -Wall -Wextra 5.9_ex-04.cpp
[arch @ voodo tc ++ pl] $ ./a.out
输入第一个整数：3
输入第二个整数：-2
i = 3 j = -2
------------------------------
x = -2 y = -2
x = 3 y = 3
---------------- --------------
使用Poiners交换
值
i = 3 j = 3
swappe dusing refrences
i = 3 j = 3
[arch @ voodo tc ++ pl] $
--- ---------------------------
x = -2 y = -2
x = 3 y = 3
------------------------------
第一行很好，但为什么它会改变x的值？在第二行。我确实
没有对x做任何事情
？
this programme compiles but runs into some strange results, a semantic-
BUG is there. i even put some "std::cout" to know what is wrong but it
does something that i did NOT order it to do:
--------------- PROGRAMME -------------------
/* Stroustrup, 5.9 exercise 4
STATEMENT:
write a programme that swaps 2 integers. solve it
using 1st with pointers and then references.
METHOD:
1.) asks the user to input 2 integers
2.) swaps the value using pointers,
prints 2 statements to show what it is doing
3.) prints the SWAPPED values
4.) swaps the vale sagain using References
5.) so after swapping 2 times we get the original input values.
*/
#include<iostream>
void swap_p(int* x, int* y);
void swap_r(int& x, int& y);
int main()
{
using std::cout;
using std::cin;
using std::endl;
int i;
int j;
cout << "Enter 1st integer: ";
cin >i;
cout << "enter 2nd integer: ";
cin >j;
cout << "i = " << i << ''\t''
<< "j = " << j << endl;
int* pi = &i;
int* pj = &j;
swap_p(pi, pj);
cout << "\nvalues swapped using Poiners\n"
<< "i = " << i << ''\t''
<< "j = " << j << endl;
swap_r(i, j);
cout << "\nswappe dusing Refrences\n"
<< "i = " << i << ''\t''
<< "j = " << j << endl;
return 0;
}
void swap_p(int* x, int* y)
{
int temp_val = *x;
x = y;
std::cout << "------------------------------\n";
std::cout << "x = " << *x << "\ty = " << *y <<''\n'';
*y = temp_val;
std::cout << "x = " << *x << "\ty = " << *y <<''\n'';
std::cout << "------------------------------\n";
}
void swap_r(int& x, int& y)
{
int temp_val = x;
x = y;
y = temp_val;
}
-------------- OUTPUT ------------------
[arch@voodo tc++pl]$ g++ -ansi -pedantic -Wall -Wextra 5.9_ex-04.cpp
[arch@voodo tc++pl]$ ./a.out
Enter 1st integer: 3
enter 2nd integer: -2
i = 3 j = -2
------------------------------
x = -2 y = -2
x = 3 y = 3
------------------------------
values swapped using Poiners
i = 3 j = 3
swappe dusing Refrences
i = 3 j = 3
[arch@voodo tc++pl]$
------------------------------
x = -2 y = -2
x = 3 y = 3
------------------------------
1st line is fine but why does it change value of "x" in 2nd line.i did
not do anything to "x"
?
推荐答案
g ++ -ansi -pedantic -Wall -Wextra 5.9_ex-04.cpp
[arch @ voodo tc ++ pl]
g++ -ansi -pedantic -Wall -Wextra 5.9_ex-04.cpp
[arch@voodo tc++pl]
./ a.out
输入第一个整数：3
输入第二个整数：-2
i = 3 j = -2
------------------------------
x = -2 y = -2
x = 3 y = 3
--------------------------- ---
使用Poiners交换
值
i = 3 j = 3
swappe dusing refrences
i = 3 j = 3
[arch @ voodo tc ++ pl]
./a.out
Enter 1st integer: 3
enter 2nd integer: -2
i = 3 j = -2
------------------------------
x = -2 y = -2
x = 3 y = 3
------------------------------
values swapped using Poiners
i = 3 j = 3
swappe dusing Refrences
i = 3 j = 3
[arch@voodo tc++pl]
---- --------------------------
x = -2 y = -2
x = 3 y = 3
------------------------------
第一行很好，但为什么它会改变x的值。在第二行。我做了什么
没有对x做任何事情
？
------------------------------
x = -2 y = -2
x = 3 y = 3
------------------------------
1st line is fine but why does it change value of "x" in 2nd line.i did
not do anything to "x"
?
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