C++ Programming - Functions - Discussion
Discussion Forum : Functions - Programs (Q.No. 21)
21.
What will be the output of the following program?
#include<iostream.h>
class IndiaBix
{
int K;
public:
void BixFunction(float, int , char);
void BixFunction(float, char, char);
};
int main()
{
IndiaBix objIB;
objIB.BixFunction(15.09, 'A', char('A' + 'A'));
return 0;
}
void IndiaBix::BixFunction(float, char y, char z)
{
K = int(z);
K = int(y);
K = y + z;
cout<< "K = " << K << endl;
}Discussion:
14 comments Page 2 of 2.
Anonymous said:
1 decade ago
Yes no issue.
If there would be a function call to the function which is only declared but not defined, then while checking the compiler would have been given a compile time error that the function is not defined.
And by the way here there is not a single definition for two declarations, rather the definition is specified only for the one function i.e.
void BixFunction(float, char, char);
If there would be a function call to the function which is only declared but not defined, then while checking the compiler would have been given a compile time error that the function is not defined.
And by the way here there is not a single definition for two declarations, rather the definition is specified only for the one function i.e.
void BixFunction(float, char, char);
Shiksha said:
1 decade ago
A single definition for two declarations isn't an issue here!
Why so?
Why so?
AKS said:
1 decade ago
char('A'+'A') is 65+65=130 which is out the signed range of char i.e., -128 to 127
So, -128+(130-128)= -126 = z
Therefore,
K = y + z = 65 + -126 = -61
So, -128+(130-128)= -126 = z
Therefore,
K = y + z = 65 + -126 = -61
(4)
Rakesh said:
1 decade ago
How this work please explain me?
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