The program will compile successfully.

Here's a step-by-step explanation of what the program does:

1. It declares a character variable `j` and assigns it the value 'a'.
2. It declares an integer variable `i` and assigns it the ASCII value of 'a' (which is 97).
3. It declares an integer variable `k` and assigns it the ASCII value of 'a' (which is 97). The explicit cast to `int` is redundant here because the conversion from `char` to `int` is implicit in C.
4. It declares an integer variable `l` and assigns it the value 75.
5. It declares a character variable `m` and assigns it the ASCII character corresponding to the integer 75 (which is 'K'). The explicit cast to `char` is necessary here to convert from `int` to `char`.
6. It prints the values of `i` and `k` (both will be 97).
7. It prints the values of `l` (which is 75) and `m` (which is 'K').
8. It waits for the user to press a key before the program ends. This is done using the `getchar()` function.
9. It returns 0, indicating that the program has ended successfully.

Question

The program will compile successfully.

Here's a step-by-step explanation of what the program does:

1. It declares a character variable `j` and assigns it the value 'a'.
2. It declares an integer variable `i` and assigns it the ASCII value of 'a' (which is 97).
3. It declares an integer variable `k` and assigns it the ASCII value of 'a' (which is 97). The explicit cast to `int` is redundant here because the conversion from `char` to `int` is implicit in C.
4. It declares an integer variable `l` and assigns it the value 75.
5. It declares a character variable `m` and assigns it the ASCII character corresponding to the integer 75 (which is 'K'). The explicit cast to `char` is necessary here to convert from `int` to `char`.
6. It prints the values of `i` and `k` (both will be 97).
7. It prints the values of `l` (which is 75) and `m` (which is 'K').
8. It waits for the user to press a key before the program ends. This is done using the `getchar()` function.
9. It returns 0, indicating that the program has ended successfully.

Knowee AI · Accepted Answer

The program will compile successfully.

Here's a step-by-step explanation of what the program does:

1. It declares a character variable `j` and assigns it the value 'a'.
2. It declares an integer variable `i` and assigns it the ASCII value of 'a' (which is 97).
3. It declares an integer variable `k` and assigns it the ASCII value of 'a' (which is 97). The explicit cast to `int` is redundant here because the conversion from `char` to `int` is implicit in C.
4. It declares an integer variable `l` and assigns it the value 75.
5. It declares a character variable `m` and assigns it the ASCII character corresponding to the integer 75 (which is 'K'). The explicit cast to `char` is necessary here to convert from `int` to `char`.
6. It prints the values of `i` and `k` (both will be 97).
7. It prints the values of `l` (which is 75) and `m` (which is 'K').
8. It waits for the user to press a key before the program ends. This is done using the `getchar()` function.
9. It returns 0, indicating that the program has ended successfully.

ablesWhat will be the status of this program upon compilation?#include <stdio.h>int main() { char j = 'a'; int i = j; int k = (int)j; int l = 75; char m = (char)l; printf("value of i = %d, k = %d\n", i, k); printf("value of l = %d, m = %c\n", l, m); getchar(); return 0;}

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