C Programming - Input / Output - Discussion
Discussion Forum : Input / Output - Yes / No Questions (Q.No. 3)
3.
Will the following program work?
#include<stdio.h>
int main()
{
int n=5;
printf("n=%*d\n", n, n);
return 0;
}
Answer: Option
Explanation:
It prints n= 5
Discussion:
22 comments Page 2 of 3.
Vijeth said:
1 decade ago
It is right justified to n. , if you use "%-*d" then it will be left justified. n can be any variable.
Shashank said:
1 decade ago
* means it specifies the width and gives those no.of spaces.
According to Nitesh's code whatever value you give to n, those many spaces are appended.
According to Nitesh's code whatever value you give to n, those many spaces are appended.
Sunitha said:
1 decade ago
Here in printf("n=%*d\n", n, n);
("%*d",n,n) tells the compiler to skip the first n value.
Thus second n value gets printed.
("%*d",n,n) tells the compiler to skip the first n value.
Thus second n value gets printed.
NITESH KHATRI said:
1 decade ago
#include<stdio.h>
int main()
{
int n=5,l=9;
printf("n=%*d\n", n, l);
return 0;
}
What @Medha said is correct, '*' will ignore the first variable and will print the value of l
Try to run this program, all confusion will be cleared.
int main()
{
int n=5,l=9;
printf("n=%*d\n", n, l);
return 0;
}
What @Medha said is correct, '*' will ignore the first variable and will print the value of l
Try to run this program, all confusion will be cleared.
Medha said:
1 decade ago
One n is ignored with 4 space because of '*'and another n value is taken.
Nandhu said:
1 decade ago
Please anybody can explain this.
Lakshmidevi said:
1 decade ago
Using * means it skips one variable, so only one n value is taken and another n value is ignored.
Veena said:
1 decade ago
I thought here two n's are not required. Please any one can explain.
Rozario selvanathan said:
1 decade ago
Anyone please make it clear...
Madhusudan said:
1 decade ago
'*' will provide 4 spaces in between 'n=' and 5.
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