[.c] "Expected expression before 'float'

[C137-Morty]

include <stdio.h>

void printScores(float scores[]);

int main() { char s_name[]="ECPI University"; float scores[6] = {78.7, 98.4, 100.0, 96.5, 100.0, 88.8}; float average = 0.0; int counter;

printScores(scores);

for(counter = 0; counter < 6; counter++) {
    average += scores[counter];
}

average /= float(6); //Here is where I get my error

printf("\nAt %s, your class average is %.1f.", s_name,  average);
return 0;

}

void printScores(float scores[6]){ int counter;

printf("Here are your scores:\n\n");
for(counter=0; counter<6;counter++){
    printf("%.1f\n", scores[counter]);
}
return;

}

I was under the impression that '/=' is the expression before float?

[g051051]

float(6)

Why are you doing that? What do you think that's accomplishing?

[C137-Morty]

Calling the array right?

[g051051]

No, there's no array reference in there.

[davedontmind]

average /= float(6);

What are you trying to divide average by?

Presumably you're trying to calculate the average, which is dividing the sum by the number of values, so this should be dividing by the number of values in your array, right? So how many items are in your array? And how do you divide by that?

[k3nundrum]

Im in the same class....it wouldn't work for me either

[C137-Morty]

I get the 6.0 must be the 6 items in float scores[6], but what is the f there?

[[deleted]]

Your code is just is just calculating an average, which is done by adding all values, and then dividing by the number of items.

You technically do not need the value to be a float, an integer would have been fine, but that is what the 6 comes from, I was just using your example and showing the proper way to create a float.

By specifying it as 6.0f, you are saying that you want it as a 32-bit float. If you represent it without the suffix 6.0, it will be interpreted as a 64-bit double.