C Programming - C Preprocessor - Discussion
Discussion Forum : C Preprocessor - Find Output of Program (Q.No. 4)
4.
What will be the output of the program?
#include<stdio.h>
#define JOIN(s1, s2) printf("%s=%s %s=%s \n", #s1, s1, #s2, s2);
int main()
{
char *str1="India";
char *str2="BIX";
JOIN(str1, str2);
return 0;
}
Discussion:
35 comments Page 3 of 4.
Saurabh Tiwari said:
1 decade ago
Yup.... arun and puja both of you are right because
#define JOIN(s1, s2) printf("%s=%s %s=%s \n", #s1, s1, #s2, s2);
Here
1) #s1 is equal to "variable name" i.e (str1) so it will print same as it is.
2) s1 that will print the "value of str1" i.e (India along with =)
and so on.
#define JOIN(s1, s2) printf("%s=%s %s=%s \n", #s1, s1, #s2, s2);
Here
1) #s1 is equal to "variable name" i.e (str1) so it will print same as it is.
2) s1 that will print the "value of str1" i.e (India along with =)
and so on.
Kundan said:
1 decade ago
I agree with Natasa. For better understanding purpose I give a example code.
#include<stdio.h>
#define res(s1,s2) printf("%s=%d %s=%d",#s1,s1,#s2,s2);
void main()
{
int a=4,b=3;
res(a,b);
}
Output of this code is
a=4 b=3
#include<stdio.h>
#define res(s1,s2) printf("%s=%d %s=%d",#s1,s1,#s2,s2);
void main()
{
int a=4,b=3;
res(a,b);
}
Output of this code is
a=4 b=3
Natasha said:
1 decade ago
# is a stringizing operator. it will convert the argument into string.
eg #define df(a) printf(#a)
main()
{
int b=7;
df(b);
}
This will give result
b
as the statement df(b) was converted to printf("b");
eg #define df(a) printf(#a)
main()
{
int b=7;
df(b);
}
This will give result
b
as the statement df(b) was converted to printf("b");
(1)
Shekar said:
1 decade ago
# operator stringrises the operand. For example #str1 is replaced by "str1" during compilation.
Pooja said:
1 decade ago
Yup.its true..firstly it is concate the 2 string nd than declared with the join s1,s2..
so,#s1 gives output as str1,#s2 gives output as str2
so.str1=india,str2=bix
so,#s1 gives output as str1,#s2 gives output as str2
so.str1=india,str2=bix
Arun said:
1 decade ago
#s1,#s2 - this will print the variable name which has been declared with the join
s1,s2 - which will display the characters
str1 and str2 are the variables and *str1 and * str2 are the pointers which has declred as char.
s1,s2 - which will display the characters
str1 and str2 are the variables and *str1 and * str2 are the pointers which has declred as char.
Vishal said:
1 decade ago
I am not getting it. Any one can explain it properly please.
Wikiok said:
1 decade ago
#define JOIN(s1, s2) printf("%s=%s %s=%s \n", #s1, s1, #s2, s2);
after preprocessing:
printf("%s=%s %s=%s \n", "str1", str1, "str2", str2);
so:
parameter -> parameter
#parameter -> "parameter"
param1##param2 -> param1param2
Or not? :-)
after preprocessing:
printf("%s=%s %s=%s \n", "str1", str1, "str2", str2);
so:
parameter -> parameter
#parameter -> "parameter"
param1##param2 -> param1param2
Or not? :-)
Vijendra said:
1 decade ago
What will be the answer the S1, S2?
Rohit said:
1 decade ago
To concatenate we use dobule hash but single can also do this
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