Assign numMatches with the number of elements in userValues that equal matchValue. userValues has NUM_VALS elements. Ex: If userValues is {2, 1, 2, 2} and matchValue is 2, then numMatches should be 3. Your code will be tested with the following values: matchValue: 2, userValues: {2, 1, 2, 2} (as in the example program above) matchValue: 0, userValues: {0, 0, 0, 0} matchValue: 10, userValues: {20, 50, 70, 100} (Notes)

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Assign `numMatches` with the number of elements in `userValues` that equal `matchValue`. `userValues` has `NUM_VALS` elements. For example, if `userValues` is {2, 1, 2, 2} and `matchValue` is 2, then `numMatches` should be 3.

Your code will be tested with the following values:
- `matchValue`: 2, `userValues`: {2, 1, 2, 2} (as in the example program above)
- `matchValue`: 0, `userValues`: {0, 0, 0, 0}
- `matchValue`: 10, `userValues`: {20, 50, 70, 100}

(Notes)

```c
#include <stdio.h>

int main(void) {
    const int NUM_VALS = 4;
    int userValues[NUM_VALS];
    int i;
    int matchValue;
    int numMatches = -99;  // Assign numMatches with 0 before your for loop

    scanf("%d", &matchValue);

    for (i = 0; i < NUM_VALS; ++i) {
        scanf("%d", &(userValues[i]));
    }

    /* Your solution goes here */

    printf("matchValue: %d, numMatches: %d\n", matchValue, numMatches);

    return 0;
}
```

- The section to add the solution is marked by a comment.
- This program is designed to read a value for `matchValue` and an array `userValues` of size `NUM_VALS`. It then calculates how many times `matchValue` appears in the `userValues` array and stores the count in `numMatches`.
- The example includes two tested scenarios with specific values for `matchValue` and `userValues`.
Transcribed Image Text:Assign `numMatches` with the number of elements in `userValues` that equal `matchValue`. `userValues` has `NUM_VALS` elements. For example, if `userValues` is {2, 1, 2, 2} and `matchValue` is 2, then `numMatches` should be 3. Your code will be tested with the following values: - `matchValue`: 2, `userValues`: {2, 1, 2, 2} (as in the example program above) - `matchValue`: 0, `userValues`: {0, 0, 0, 0} - `matchValue`: 10, `userValues`: {20, 50, 70, 100} (Notes) ```c #include <stdio.h> int main(void) { const int NUM_VALS = 4; int userValues[NUM_VALS]; int i; int matchValue; int numMatches = -99; // Assign numMatches with 0 before your for loop scanf("%d", &matchValue); for (i = 0; i < NUM_VALS; ++i) { scanf("%d", &(userValues[i])); } /* Your solution goes here */ printf("matchValue: %d, numMatches: %d\n", matchValue, numMatches); return 0; } ``` - The section to add the solution is marked by a comment. - This program is designed to read a value for `matchValue` and an array `userValues` of size `NUM_VALS`. It then calculates how many times `matchValue` appears in the `userValues` array and stores the count in `numMatches`. - The example includes two tested scenarios with specific values for `matchValue` and `userValues`.
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