没有得到这个程序的预期输出 [英] Not getting the expected output in this program
问题描述
大家好，
新手在这里，PLZ帮助.......
这就是我编码的（它非常非常很简单，我仍然面临着这段代码的问题）--------->
__________________________________________________ _______________
Hello everyone,
Newbie here, PLZ HELP.......
This is what I coded (it''s very very simple, still I''m facing problem with this code) --------->
__________________________________________________ _______________
推荐答案
大家好，
新手在这里，PLZ HELP ..... ..
这是我编码的（它非常简单，我仍然面临着问题）他的代码）--------->
__________________________________________________ _______________
#include< conio.h>
#include< stdio.h>
int main（）
{
int i = 0;
int j = 1 ;
clrscr（）;
printf（" \ n％d"，i ++&& ++ j）;
printf（" \ n％d％d"，i，j）;
getch（）;
返回（0）;
}
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**************
**输出**
**************
0
1 1
__________________________________________________ _______________
我正在使用Borland Turbo C ++ 3.0。
我希望代码
printf （\ n％d，i ++&& ++ j）;
将j递增1并输出0然后将i递增1.
现在我变为1，j变为2.
现在执行以下代码时
printf（" \ n％d％d"， i，j）;
我现在希望将i和j的值打印为1和2的代码，但我输出为
1 1.
我哪里错了，请指导我。
__________________________________________________ _______________
感谢E有人提前...............
Hello everyone,
Newbie here, PLZ HELP.......
This is what I coded (it''s very very simple, still I''m facing problem with this code) --------->
__________________________________________________ _______________
#include<conio.h>
#include<stdio.h>
int main()
{
int i=0;
int j=1;
clrscr();
printf("\n %d",i++ && ++j);
printf("\n %d %d",i,j);
getch();
return(0);
}
__________________________________________________ _______________
**************
**OUTPUT**
**************
0
1 1
__________________________________________________ _______________
I''m using Borland Turbo C++ 3.0.
I expect the code
printf("\n %d",i++ && ++ j);
to increment j by 1 and output 0 and then increment i by 1.
Now i becomes 1 and j becomes 2.
Now when the code below is executed
printf("\n %d %d",i,j);
I expect now that the code to print value of i and j as 1 and 2, but I get output as
1 1.
Where am I wrong, Please guide me.
__________________________________________________ _______________
THANKS TO EVERYONE IN ADVANCE...............
短路评估会导致问题。短路评估意味着只要表达式的结果已知，表达式的其余部分就不会被评估。
在任何条件下 a&& b ，如果a为假，则无论b的值如何，表达式都不能为真。所以表达式的其余部分被跳过，因为我们已经知道了答案。
当你做 i ++&& ++ j ，我从零开始，因为你使用了后缀增量运算符，直到分号才增加i。 0 = false，因此递增j的其余语句永远不会执行。
有你的解释，但更重要的是要意识到：
- 在函数调用中混合布尔表达式通常是一个坏主意，并且
- 在更复杂的语句中使用一元运算符通常也是一个坏主意。
两者都会导致意外的结果和难以阅读的代码。
Short-circuit evaluation causes the problem. Short-circuit evaluation means as soon as the result of an expression is known, the rest of the expression is not evaluated.
In any condition a && b, if a is false then the expression cannot be true regardless of the value of b. So the rest of the expression is skipped because we know the answer already.
When you do i++ && ++j, i starts at zero and because you used the postfix increment operator, i is not incremented until the semicolon. 0 = false, thus the rest of the statement incrementing j is never executed.
There''s your explanation, but what''s more important to realize is that:
- mixing boolean expressions inside function calls is generally a bad idea, and
- using unary operators within more complex statements is also generally a bad idea.
Both will often lead to unexpected results and hard-to-read code.
短路评估会导致问题。短路评估意味着只要表达式的结果已知，表达式的其余部分就不会被评估。
在任何条件下 a&& b ，如果a为假，则无论b的值如何，表达式都不能为真。所以表达式的其余部分被跳过，因为我们已经知道了答案。
当你做 i ++&& ++ j ，我从零开始，因为你使用了后缀增量运算符，直到分号才增加i。 0 = false，因此递增j的其余语句永远不会执行。
有你的解释，但更重要的是要意识到：
- 在函数调用中混合布尔表达式通常是一个坏主意，并且
- 在更复杂的语句中使用一元运算符通常也是一个坏主意。
两者都会导致意外的结果和难以阅读的代码。
Short-circuit evaluation causes the problem. Short-circuit evaluation means as soon as the result of an expression is known, the rest of the expression is not evaluated.
In any condition a && b, if a is false then the expression cannot be true regardless of the value of b. So the rest of the expression is skipped because we know the answer already.
When you do i++ && ++j, i starts at zero and because you used the postfix increment operator, i is not incremented until the semicolon. 0 = false, thus the rest of the statement incrementing j is never executed.
There''s your explanation, but what''s more important to realize is that:
- mixing boolean expressions inside function calls is generally a bad idea, and
- using unary operators within more complex statements is also generally a bad idea.
Both will often lead to unexpected results and hard-to-read code.
谢谢
这就是我想要的..........
我知道SHORT-CIRCUIT评估，但我不知道我会在这里面对它。
请再告诉我一件事。术语短路评估是指短路评估吗？是C中的技术术语还是这个现象的随机名称？
再次感谢，非常感谢......
THANKS
That''s what I was lokking for..........
I knew about SHORT-CIRCUIT evaluation, but I didn''t knew that I will face it here.
Please tell me one more thing. Does the term "SHORT-CIRCUIT evaluation" is technical terminology in C or it''s just some random name for this phenomenon?
ONCE AGAIN, THANK YOU VERY MUCH......
短路评估会导致问题。短路评估意味着只要表达式的结果已知，表达式的其余部分就不会被评估。
在任何条件下 a&& b ，如果a为假，则无论b的值如何，表达式都不能为真。所以表达式的其余部分被跳过，因为我们已经知道了答案。
当你做 i ++&& ++ j ，我从零开始，因为你使用了后缀增量运算符，直到分号才增加i。 0 = false，因此递增j的其余语句永远不会执行。
有你的解释，但更重要的是要意识到：
- 在函数调用中混合布尔表达式通常是一个坏主意，并且
- 在更复杂的语句中使用一元运算符通常也是一个坏主意。
两者都会导致意外的结果和难以阅读的代码。
Short-circuit evaluation causes the problem. Short-circuit evaluation means as soon as the result of an expression is known, the rest of the expression is not evaluated.
In any condition a && b, if a is false then the expression cannot be true regardless of the value of b. So the rest of the expression is skipped because we know the answer already.
When you do i++ && ++j, i starts at zero and because you used the postfix increment operator, i is not incremented until the semicolon. 0 = false, thus the rest of the statement incrementing j is never executed.
There''s your explanation, but what''s more important to realize is that:
- mixing boolean expressions inside function calls is generally a bad idea, and
- using unary operators within more complex statements is also generally a bad idea.
Both will often lead to unexpected results and hard-to-read code.
__________________________________________________ _______________
帮助，新手在这里--------->
我仍感到困惑......
正如你所说，如果左操作数的计算结果为FALSE，则不会计算逻辑运算符（&&）右侧的操作数。但是根据C语言的优先级规则，UNARY运算符（例如这个问题中的++）比二元运算符具有更高的优先级。
因此，无论运算符左侧的表达式值是什么？首先，必须单独评估i ++和++ j，因为它们的优先级高于&&运营商。
我知道我错了（基本概念），所以请指导我.........
感谢所有人提前.........
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HELP, newbie here--------->
Still I am confused......
As you said, the operand on the right side of logical operator (&&) is not evaluated if left operand evaluates to FALSE. But according to priority rules of C language, UNARY operators (for example ++ in this question) have a higher priority then binary operators.
Therefore regardless of value of expression on left of operator &&, first of all i++ and ++j must be evaluated separately as they have a higher priority than && operator.
I know I''m wrong somewhere (in basic concept), so please guide me.........
THANKS TO EVERYONE IN ADVANCE.........
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