@Ashima..
u r right...it is always printing first float variable value
On Sat, Sep 17, 2011 at 2:05 PM, Ashima . wrote:
> @yogesh : but if in place of x=1 in above code, I write t=54. that means i
> m assigning somevalue to a float variable.Then it should print 54
> afterwards.
> but its ag
@yogesh : but if in place of x=1 in above code, I write t=54. that means i
m assigning somevalue to a float variable.Then it should print 54
afterwards.
but its again printing 98. how does it go with ur concept.
Ashima
M.Sc.(Tech)Information Systems
4th year
BITS Pilani
Rajasthan
On Fri, Sep 1
@yogesh :ya u right.
On Sat, Sep 17, 2011 at 11:58 AM, sukran dhawan wrote:
> x== 9 is not an error !!! it simply returns a non zero value if its true
> and 0 if its false
>
> On Sat, Sep 17, 2011 at 11:39 AM, praveen raj wrote:
>
>> This will show syntax error due to x==9 and otherwise since
x== 9 is not an error !!! it simply returns a non zero value if its true and
0 if its false
On Sat, Sep 17, 2011 at 11:39 AM, praveen raj wrote:
> This will show syntax error due to x==9 and otherwise since memory address
> given at the declarAtion there fore x value will. Change at every
> ass
x==9 is not an error...
it will be 1 when true and 0 when false...and both are executable
statements...
when we print float with %d then compiler prints 0...and when we print
integer with %f then compiler prints last assigned or printed float value to
any float variablehere it is assigned for
This will show syntax error due to x==9 and otherwise since memory address
given at the declarAtion there fore x value will. Change at every
assignment , but i m not sure printf will give an error or not . Plz reply.
On 17-Sep-2011 1:13 AM, "Anshul AGARWAL" wrote:
>
> #include
> int main()
> {fl
Printing a floating point with %d conversion specifier will not print the
expected resultss !
On Sat, Sep 17, 2011 at 1:13 AM, Anshul AGARWAL
wrote:
> #include
> int main()
> {float t;
> long x;
>
>
> t=98;
> printf("%d\n",t);
> printf("%f\n",x);
> {
> x=1;
>
#include
int main()
{float t;
long x;
t=98;
printf("%d\n",t);
printf("%f\n",x);
{
x=1;
printf("%f\n",x);
{
x=30;
printf("%f\n",x);
}
printf("%f\n",x);
}
x==9;
printf("%f\n",x);
}
-
char type is signed by default, so it represents value 255 as -1.
Code outputs first byte as unsigned int value, where -1 == 4294967295.
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2.
#include
using namespace std;
int main(){
int n = 255, i;
char *p = (char *)&n;
for(i = 0; i < sizeof(int); i++)
cout<<(unsigned int)p[i]<<"\n";
cin.get();
return 0;
}
Output:
4294967295
0
0
0
I think p[0] is .So why it is giving 4294967295 in the first line.
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# include
#include
#define concatinate(a,b) a##b
#define same1(a) #a
#define same2(a) same1(a)
int main(){
printf("%s\n",same2(concatinate(1,2)));
printf("%s\n",same1(concatinate(1,2)));
getch();
return 0;
}
Output is:
12
concatinate(1,2)
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