//Please don't copy-paste this program ; make your own implementation
#include <cstdio>
#include <iostream>
using namespace std;
int main()
{
int a;
a = 5;
cout << ++a << endl; // pre-increment -- 6
cout << --a << endl; // pre-decrement -- 4
cout << a++ << endl; // post-incremenr -- 5
cout << a-- << endl; // post-decrement -- 5
cout << a++ << a++ << endl; // 5 -(+1)-> 6
cout << a++ + a++ + a++ << endl ; // 5 -(+1)-> 6 -(+1)-> 7 = 18
//This goes left-to-right
cout << a++ + a++ + --a; // 5 -(+1)-> 6 -(+1)-> 7 (-1) -->6 =17
// it goes right-to-left
cout << a++ , a++ , a++; //7, <-(+1)- 6, <-(+1)- 5
//it goes 1st right-to-left and then left-to-right
cout << a++ + a++ + a--, ++a; // 21,6
printf ( "%d %d %d %d %d\n",x++, x--, ++x, --x, x++); // 5 6 6 5 5
//For Practice
int b = 5, c = 5, i, j, j1;
i = a++ + a++ + a++;
j = ++b + ++b + ++b;
j1 = c++ + c++ + c--;
printf ("\n value of i using printf:%d\n", i);
cout << "\n Value of i : " << i << endl
<< "\n Value of j : " << j << endl
<< "\n Value of j1 : " << j1 << endl;
printf("value of 1st print is: %d\n", 59);
printf("value of 2nd print is: %c\n\n", 59);
}
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//What is the output of this program
#include <stdio.h>
int main()
{
int x=20,y=35;
x=x++ + y++;
y=++x + ++y;
printf ( "%d, %d", x,y);
return 0;
}
The output for your program is 57, 94.
Explanation:
x = 20, y = 35
x = x++ + y++ = 20 + 35 (x and y would be incremented after the addition) = 55
x++ -----> x = 55 + 1 = 56
y++ -----> y = 35 + 1 = 36
++x -----> x = 56 + 1 = 57 (x and y would be incremented
++y -----> y = 36 + 1 = 37 before the addition)
y = ++x + ++y = 57 + 37 = 94
Hence final values of x and y are 57 and 94 respectively.
Understand this way:
d=x++; //the value of x is stored to d and d is increment by 1.x remains the same.d=20 x=21
printf ("%d %d",d,x); // 20 21
x = x++;
printf("%d",x); //21
d = x++ + 1;
x = x++ + 1;
printf("%d %d"); //d=21 x=22
d=++x; //here increment is done first after that value is stored=x=21
For further detail click here.
#include <cstdio>
#include <iostream>
using namespace std;
int main()
{
int a;
a = 5;
cout << ++a << endl; // pre-increment -- 6
cout << --a << endl; // pre-decrement -- 4
cout << a++ << endl; // post-incremenr -- 5
cout << a-- << endl; // post-decrement -- 5
cout << a++ << a++ << endl; // 5 -(+1)-> 6
cout << a++ + a++ + a++ << endl ; // 5 -(+1)-> 6 -(+1)-> 7 = 18
//This goes left-to-right
cout << a++ + a++ + --a; // 5 -(+1)-> 6 -(+1)-> 7 (-1) -->6 =17
// it goes right-to-left
cout << a++ , a++ , a++; //7, <-(+1)- 6, <-(+1)- 5
//it goes 1st right-to-left and then left-to-right
cout << a++ + a++ + a--, ++a; // 21,6
printf ( "%d %d %d %d %d\n",x++, x--, ++x, --x, x++); // 5 6 6 5 5
//For Practice
int b = 5, c = 5, i, j, j1;
i = a++ + a++ + a++;
j = ++b + ++b + ++b;
j1 = c++ + c++ + c--;
printf ("\n value of i using printf:%d\n", i);
cout << "\n Value of i : " << i << endl
<< "\n Value of j : " << j << endl
<< "\n Value of j1 : " << j1 << endl;
printf("value of 1st print is: %d\n", 59);
printf("value of 2nd print is: %c\n\n", 59);
}
-----------------------------------------------------------------------------------------------
//What is the output of this program
#include <stdio.h>
int main()
{
int x=20,y=35;
x=x++ + y++;
y=++x + ++y;
printf ( "%d, %d", x,y);
return 0;
}
The output for your program is 57, 94.
Explanation:
x = 20, y = 35
x = x++ + y++ = 20 + 35 (x and y would be incremented after the addition) = 55
x++ -----> x = 55 + 1 = 56
y++ -----> y = 35 + 1 = 36
++x -----> x = 56 + 1 = 57 (x and y would be incremented
++y -----> y = 36 + 1 = 37 before the addition)
y = ++x + ++y = 57 + 37 = 94
Hence final values of x and y are 57 and 94 respectively.
Understand this way:
d=x++; //the value of x is stored to d and d is increment by 1.x remains the same.d=20 x=21
printf ("%d %d",d,x); // 20 21
x = x++;
printf("%d",x); //21
d = x++ + 1;
x = x++ + 1;
printf("%d %d"); //d=21 x=22
d=++x; //here increment is done first after that value is stored=x=21
For further detail click here.
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