+1 to yogesh
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4 and 5. size of will check its first operand. so no evaluation takes
place...
On Sat, Sep 17, 2011 at 2:32 PM, abhinav gupta wrote:
> Ans will be:4,5
>
>
> On Sat, Sep 17, 2011 at 2:23 PM, Sanjay Rajpal wrote:
>
>> #include
>>
>> int main()
>> {
>> int a=5;
>> printf("Size : %d\n",sizeof( a
4,5
statements inside sizeof() does not get executed...it will tell just size...
int a=3;
printf("%d",sizeof(a++));
here a++ will not be executed ...it will just tell size ...
Also, lets suppose it execute statements ...then sizeof(int),sizeof(node *)
...will always produce error because int a
4,7..if we assume the sizeof(int) as 4 bytes..
On Sat, Sep 17, 2011 at 2:32 PM, abhinav gupta wrote:
> Ans will be:4,5
>
>
> On Sat, Sep 17, 2011 at 2:23 PM, Sanjay Rajpal wrote:
>
>> #include
>> int main()
>> {
>> int a=5;
>> printf("Size : %d\n",sizeof( a =15/2));
>> printf("A is %d.",a);
Ans will be:4,5
On Sat, Sep 17, 2011 at 2:23 PM, Sanjay Rajpal wrote:
> #include
>
> int main()
> {
> int a=5;
> printf("Size : %d\n",sizeof( a =15/2));
> printf("A is %d.",a);
> }
>
> What will be the value of a now ? Plz explain.
> Sanju
> :)
>
> --
> You received this message because you
@priya:
compiler thinks that " a++; " statement is correct... But for that to do...'
a' should be of type modifiable ... so it gives error saying that "If u want
to modify , that should be lvalue, so lvalue required"
On Sun, Sep 4, 2011 at 2:27 AM, Yogesh Yadav wrote:
> @Priya:
> Because a
@Priya:
Because a is *unmodifiable* LValue. When we are using a=a+1(a++) , this
means that we are changing the *base address* of array a[], and that can not
be possible due to Library files.
On Sun, Sep 4, 2011 at 11:49 AM, priya ramesh <
love.for.programm...@gmail.com> wrote:
> @yogesh: Thanks
@yogesh: Thanks a lot for the explaination. Good one indeed. Can you plz
tell me y the compiler says lvalue required for the statement
a++ when a is an lvalue??
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@Priya:
Lvalue is not a contant value. It can be modifiable or unmodifiable. Lvalue
stands for location value that means that we can store some value here in
this location and rvalue stands for read value that means that it can be
stored at some location that is in memory having lvalue.
Here in c
if it is an l value, the compiler should say
"cannot modify const value"
instead it says
"l value required"
This statement proves that a is an r value in main. right??
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@priya :array is a constant pointer to a non constant data remember
On Sat, Sep 3, 2011 at 6:48 PM, priya ramesh wrote:
> ok but y does the compiler say lvalue required if you perform a++??
>
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lvalue
On Sat, Sep 3, 2011 at 6:44 PM, himanshu kansal wrote:
> an array is always an *unmodifiable* l-value.
>
>
> On Sat, Sep 3, 2011 at 6:23 PM, priya ramesh <
> love.for.programm...@gmail.com> wrote:
>
>> In main, any array say char a[200],
>> a refers to an r value or an lvalue??
>>
ok but y does the compiler say lvalue required if you perform a++??
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an array is always an *unmodifiable* l-value.
On Sat, Sep 3, 2011 at 6:23 PM, priya ramesh wrote:
> In main, any array say char a[200],
> a refers to an r value or an lvalue??
>
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>
@harshalthanks , can u tell me how to find what features are compiler
dependent
On Wed, Jun 8, 2011 at 4:28 PM, Harshal wrote:
> This depends upon the implementation. Hence, your program is not
> portable. It's allowed to return NULL, and it's allowed to return a non-NULL
> pointer. Both wa
This depends upon the implementation. Hence, your program is not
portable. It's allowed to return NULL, and it's allowed to return a non-NULL
pointer. Both ways are sanctioned by the Standard.
On Wed, Jun 8, 2011 at 4:20 PM, coder dumca wrote:
> can any one tell me why this code perfectly runnin
@balaji: that's the default value for the parameter. Basically, the
advantage of the default parameter is that you need not to explicitly define
the non-parameterized constructor.
See the difference yourself with following 2 examples.
1) *class A {
int m;
A (int a
Actually
A(int *m=0*)
> {
> a=m;
> }
> not
> A*(int m*) {
>a = m;
> }
>
means m has a default value of 0 ie this value will be used if no parameter
is given . So when you pass it a parameter default value is simply ignored.
On Wed, May 25, 2011 at 12:15 PM, Bal
ya.. thanks :) it works. but.. we are initializing m to 0 in everycall ryt..
? then how does 1,2,3,is initialized??
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Change it to that.. It will still work.. Don't worry :)
On Tue, May 24, 2011 at 10:56 PM, Balaji S wrote:
> but
> constructor is
>
>
> A(int *m=0*)
> {
> a=m;
> }
> not
>
> A*(int m*) {
>a = m;
> }
>
> ???
>
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but
constructor is
A(int *m=0*)
{
a=m;
}
not
A*(int m*) {
a = m;
}
???
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This way you can do:
#include
>
> using namespace std;
>
> class A {
> public:
> int a;
>
> A(int m) {
> a = m;
> }
>
> A() {
> }
> };
>
> int main()
> {
> int i;
>
> A
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