整数int i; * i ++和++ *我在增量后有一个不同的整数值 [英] integer int i; *i++ and ++*i have a different integer value after the increment
问题描述
大家好，
我有一个值为10的整数指针，但它会在预增量和后增量后返回不同的
返回值。 br />
如果有人可以解释这个
行为，我将非常感激。
下面列出的是程序和输出 - 在vc ++ 6中执行。
int main（int argc，char * argv []）
{
int * i =（int *）calloc（sizeof（int），1）;
* i = 10;
cout<< " * i ++ << * i ++<< endl;
cout<< " * i << * i<< endl;
* i = 10;
cout<< " ++ * i << ++ * i<< endl;
cout<< " * i << * i<<结束;
返回0;
}
* i ++ 10
* i -33686019
++ * i 11
* i 11
按任意键继续
Hi All,
I have an integer pointer with value 10, but it returns a different
return value after the preincrement and post increment.
I will be very thankful if somebody can give an explanation for this
behavior.
Listed below is the program and output - executed in vc++6.
int main(int argc, char* argv[])
{
int *i = (int*) calloc(sizeof(int),1);
*i = 10;
cout << " *i++ " << *i++ << endl;
cout << " *i " << *i << endl;
*i = 10;
cout << " ++*i " << ++*i << endl;
cout << " *i " << *i << endl;
return 0;
}
*i++ 10
*i -33686019
++*i 11
*i 11
Press any key to continue
推荐答案
查看运算符优先级。 operator ++的优先级高于
一元运算符*，所以
* i ++
相当于
*（i ++ ）
并不等于
（* i）++
这是你想要的。当您取消引用指针
并找到意外的大幅度值时，通常意味着
指针没有指向您认为的位置。 />
卢克
Review operator precedence. operator++ has higher precedence than
unary operator*, so
*i++
is equivalent to
*(i++)
and not equivalent to
(*i)++
which is what your presumably wanted. When you dereference a pointer
and find an unexpected, large-magnitude value, it generally means that
the pointer isn''t pointing where you thought it was.
Luke
罗本写道：
Robben wrote:
你好全部，
我有一个值为10的整数指针，但是在预增量和后增量之后返回一个不同的
返回值。
我将非常感激如果有人可以对这个行为给出解释。
下面列出的是程序和输出 - 在vc ++ 6中执行。
int main（int argc ，char * argv []）
{* / int * i =（int *）calloc（sizeof（int），1）;
不要：
int * i = new int;
* i = 10;
cout<< " * i ++ << * i ++<< ENDL;
这是*（i ++）：递增指针，返回旧地址和
取消引用它。
cout< < " * i << * i<<结束;
* i = 10;
非法，我不再指向有效记忆了。
cout<< " ++ * i << ++ * i<< ENDL;
这是++（* i）：取消引用指针并递增值。
cout<< " * i << * i<< ENDL;
别忘了删除int：
删除i;
返回0;
}
* i ++ 10
* i -33686019
++ * i 11
* i 11
Hi All,
I have an integer pointer with value 10, but it returns a different
return value after the preincrement and post increment.
I will be very thankful if somebody can give an explanation for this
behavior.
Listed below is the program and output - executed in vc++6.
int main(int argc, char* argv[])
{
int *i = (int*) calloc(sizeof(int),1);
Don''t:
int* i = new int;
*i = 10;
cout << " *i++ " << *i++ << endl;
This is *(i++) : increment the pointer, return the old address and
dereference it.
cout << " *i " << *i << endl;
*i = 10;
illegal, i does not point to valid memory anymore.
cout << " ++*i " << ++*i << endl;
and this is ++(*i) : dereference the pointer and increment the value.
cout << " *i " << *i << endl;
Don''t forget to delete the int:
delete i;
return 0;
}
*i++ 10
*i -33686019
++*i 11
*i 11
Jonathan
Jonathan
>不要忘记删除int：
> Don''t forget to delete the int:
删除i;
delete i;
记住当然不要混合使用;新"或删除单独使用
C风格（de）分配。但既然我同意你的建议
使用new而不是calloc，没问题。
Luke
Remembering, of course, not to mix "new" or "delete" in isolation with
C-style (de)allocation. But since I agree with your recommendation to
use new rather than calloc, no problem.
Luke
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