r*******y
发帖数: 1081 | 1 I use g++ compiler. Something strange here.
case 1:
int a[3]={1, 10, 100};
int *p = a;
cout << *p++ << endl << *p <
The out put is 1 1
I can not understand this. It is supposed to be 1 10, right?
case 2:
if I write it as
cout << *p++ <
cout << *p <
the output is 1 10 as it is supposed to be.
case 3:
Also if I have this code
int i = 1;
cout << i++ <
The output is also as supposed to be, that is 1 2;
So here case 2 and case 3 are well understood. What is wrong with case 1?
Thanks. | p***o
发帖数: 1252 | 2 1 and 3 are undefined. Google "sequence point".
【在 r*******y 的大作中提到】
: I use g++ compiler. Something strange here.
: case 1:
: int a[3]={1, 10, 100};
: int *p = a;
: cout << *p++ << endl << *p <: The out put is 1 1
: I can not understand this. It is supposed to be 1 10, right?
: case 2:
: if I write it as
: cout << *p++ <
| r*******y
发帖数: 1081 | 3 thanks a lot.
【在 p***o 的大作中提到】
: 1 and 3 are undefined. Google "sequence point".
| f**********w
发帖数: 93 | 4 from the twiki "An often-cited example is the expression i=i++, which both
assigns i to itself and increments i. The final value of i is ambiguous,
because, depending on the language semantics, the increment may occur before
, after or interleaved with the assignment. The definition of a particular
language might specify one of the possible behaviors or simply say the
behavior is undefined. In C and C++, evaluating such an expression yields
undefined behavior."
Is this true? i=i++ should be expanded as i.operator=(i.operator++(int) ).
Is this behavior undefined? |
|
