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Consider the following C program:
#include <stdio.h>
int jumble(int x, int y){
x=2*x+y;
return x; }
int main(){
int x=2, y=5;
y= jumble(y,x);
x= jumble(y,x);
printf("%d \n", x);
return 0;
}
The value printed by the program is _________
    Correct answer is '26'. Can you explain this answer?
    Verified Answer
    Consider the following C program:#include <stdio.h>int jumble(in...
    Initial value of x and y is given as 2 and 5
    y= jumble(y,x);
    In function
    5 is stored in x and 1 stored in y
    x = 2*x+y is equal to 2*5+2 = 12 x = 12
    return value update y to 12
    x= jumble(y,x);
    now 12 is stored in x and 2 stored in y
    x = 2*x+y is equal to 2*12+2 = 26
    x = 26
    return value of update x to 26 value printed is 26
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    Most Upvoted Answer
    Consider the following C program:#include <stdio.h>int jumble(in...

    Explanation:

    - Function jumble:
    - The function jumble takes two integer parameters x and y.
    - Inside the function, x is modified by multiplying it by 2 and adding y to it.
    - The modified value of x is then returned.

    - Main function:
    - In the main function, two integer variables x and y are declared and initialized with values 2 and 5 respectively.
    - The jumble function is called with parameters y and x, and the return value is assigned to y.
    - The jumble function is called again with parameters y and x, and the return value is assigned to x.
    - Finally, the value of x is printed using printf.

    - Execution flow:
    - First call to jumble: y = 2*y + x = 2*5 + 2 = 12
    - Second call to jumble: x = 2*y + x = 2*12 + 2 = 26
    - The final value of x is 26, which is printed by the program.
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    Consider the following C program:#include <stdio.h>int jumble(int x, int y){x=2*x+y;return x; }int main(){int x=2, y=5;y= jumble(y,x);x= jumble(y,x);printf("%d \n", x);return 0;}The value printed by the program is _________Correct answer is '26'. Can you explain this answer?
    Question Description
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