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what is the difference between %d and %*d in c languaga?
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Answer / sivakumar
anand and manini your expectations are wrong. because %d
give the original value of the variable and %*d give the
address of the variable.
eg:-
int a=10,b=20;
printf("%d%d",a,b);
printf("%*d%*d",a,b);
result is 10 20 1775 1775
here 1775 is the starting address of the memory allocation
for the integer.a and b having same address because of
contagious memory allocation.
| Is This Answer Correct ? | 35 Yes | 19 No |
Answer / azad sable,chiplun.
In first case i.e. '%d' the '%' indicates that the
conversion specification follows. And 'd' known as data
type charactor indicates that the no. to be read is in
intiger mode.
* is an input field which specifie field width.
example
scanf("%d%*d%d",&a,&b);
will assign the data 123 456 789 as follows.123 to a 456
skipp because of * 789 to b.
| Is This Answer Correct ? | 9 Yes | 5 No |
Answer / vishal pandey
int v=23,d=89;
printf("%d %*d",v,d);
then o/p v=23 and d=address value and address value change with processor but original value does not change.
| Is This Answer Correct ? | 3 Yes | 2 No |
Answer / shivam chaurasia
The %*d in a printf allows you to use a variable to control the field width, along the lines of:
int wid = 4;
printf ("%*d
", wid, 42);
output,...
..42
if the form is like this...
printf ("%*d %*d
", a, b);
is undefined behaviour as per the standard, since you should be providing four arguments after the format string, not two (and good compilers like gcc will tell you about this if you bump up the warning level). From C11 7.20.6 Formatted input/output functions:
If there are insufficient arguments for the format, the behavior is undefined.
It should be something like:
printf ("%*d %*d
", 4, a, 4, b);
check this link for extra detail....
http://www.cplusplus.com/reference/cstdio/printf/
| Is This Answer Correct ? | 1 Yes | 0 No |
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