C Programming - Expressions - Discussion
Discussion Forum : Expressions - Find Output of Program (Q.No. 12)
12.
What will be the output of the program?
#include<stdio.h>
int main()
{
int i=2;
printf("%d, %d\n", ++i, ++i);
return 0;
}
Answer: Option
Explanation:
The order of evaluation of arguments passed to a function call is unspecified.
Anyhow, we consider ++i, ++i are Right-to-Left associativity. The output of the program is 4, 3.
In TurboC, the output will be 4, 3.
In GCC, the output will be 4, 4.
Discussion:
43 comments Page 1 of 5.
Vishwa said:
1 decade ago
Can anybody explain this...?
Rachit said:
1 decade ago
Will the basic rule chan of post/pre increment/decrement in GCC ?
Pratik said:
1 decade ago
Please elabortate it.
Pruthvi said:
1 decade ago
Is associativity considered in the printf function also or only while evaluating an expression we consider the associativity ?
Can any one explain it. ?
Can any one explain it. ?
Kamal said:
1 decade ago
printf("%d, %d", ++i, ++i)
Here arguments in fxn prnitf() are evaluated from right to left. It is a rule in C. And at time of printing they are displayed in the Left to Right basis.
Here arguments in fxn prnitf() are evaluated from right to left. It is a rule in C. And at time of printing they are displayed in the Left to Right basis.
Pooja said:
1 decade ago
Please explain it clearly with examples.
Neethu said:
1 decade ago
printf("%d, %d", ++i, ++i)
Here arguments in fxn prnitf() are evaluated from right to left. It is a rule in C. And at time of printing they are displayed in the Left to Right basis.
Here arguments in fxn prnitf() are evaluated from right to left. It is a rule in C. And at time of printing they are displayed in the Left to Right basis.
Ravikiran said:
1 decade ago
It is right.. for example:
#include<stdio.h>
int main()
{
int a=35;
printf("%d %d %d ",a==40,a<=50,a=30);
return 0;
}
will u see the out put in this .. the out put will be
0 0 30
because it can read right to left in print statement, when it we write the options continuously(a==40,a<=50,a=30).
#include<stdio.h>
int main()
{
int a=35;
printf("%d %d %d ",a==40,a<=50,a=30);
return 0;
}
will u see the out put in this .. the out put will be
0 0 30
because it can read right to left in print statement, when it we write the options continuously(a==40,a<=50,a=30).
Yami said:
1 decade ago
#include<stdio.h>
int main()
{
int a=35;
printf("%d %d %d ",a==40,a<=50,a=30);
return 0;
}
Please explain it clearly ....
o/p : 0 0 30
int main()
{
int a=35;
printf("%d %d %d ",a==40,a<=50,a=30);
return 0;
}
Please explain it clearly ....
o/p : 0 0 30
P.Vijay kumar said:
1 decade ago
Hi Friends...
The evalution order in c is:Right to left but in c++ is Left to Right.
During the display, the values are dispaly from left to right.(both c and c++)
#include<stdio.h>
int main()
{
int a=35;
printf("%d %d %d ",a==40,a<=50,a=30);
return 0;
}
O/P : 0 1 30
For eg::
main()
{
int i=5;
printf("%d %d",++i,++i);
}
O/P:: in c::7,6 but in c++ 6,7
But it varies depends on the compiler..
The evalution order in c is:Right to left but in c++ is Left to Right.
During the display, the values are dispaly from left to right.(both c and c++)
#include<stdio.h>
int main()
{
int a=35;
printf("%d %d %d ",a==40,a<=50,a=30);
return 0;
}
O/P : 0 1 30
For eg::
main()
{
int i=5;
printf("%d %d",++i,++i);
}
O/P:: in c::7,6 but in c++ 6,7
But it varies depends on the compiler..
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