Help on determing issues with "expected identifier errors".

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My task is to create a

Creating an average_variance function to return the average of an array of data and also calculate the variance and return it in a pointer. I've created one average_variance function that I must abide by and created a for loop to calculate the average and the variance. 

Yet I get an immense amount of errors such as;
<pre>Pointers.c:10:5: error: expected identifier or ‘(’ before ‘return’
   10 |     return (sqDiff / numdata);
      |     ^~~~~~
Pointers.c:13:2: error: expected identifier or ‘(’ before ‘for’
   13 |  for (int i = 0; i<numdata; i++)
      |  ^~~
Pointers.c:13:19: error: expected ‘=’, ‘,’, ‘;’, ‘asm’ or ‘__attribute__’ before ‘<’ token
   13 |  for (int i = 0; i<numdata; i++)
      |                   ^
Pointers.c:13:30: error: expected ‘=’, ‘,’, ‘;’, ‘asm’ or ‘__attribute__’ before ‘++’ token
   13 |  for (int i = 0; i<numdata; i++)
      |                              ^~
Pointers.c:15:2: error: expected identifier or ‘(’ before ‘}’ token
   15 |  }
      |  ^
Pointers.c:16:2: error: expected identifier or ‘(’ before ‘return’
   16 |  return (sum / numdata);
      |  ^~~~~~
Pointers.c:17:1: error: expected identifier or ‘(’ before ‘}’ token
   17 | }
      | ^
Pointers.c: In function ‘main’:
Pointers.c:24:18: error: too many arguments to function ‘average_variance’
   24 |     double avg = average_variance(values, 6, & variance);
      |                  ^~~~~~~~~~~~~~~~
Pointers.c:2:8: note: declared here
    2 | double average_variance(double data[], int numdata)
      |        ^~~~~~~~~~~~~~~~
Pointers.c:25:12: error: stray ‘\342’ in program
   25 |     printf("average = % f variance = % f\ n", average, variance);
      |            ^
Pointers.c:25:15: error: ‘average’ undeclared (first use in this function)
   25 |     printf("average = % f variance = % f\ n", average, variance);
      |             ^~~~~~~
Pointers.c:25:15: note: each undeclared identifier is reported only once for each function it appears in
Pointers.c:25:25: error: expected expression before ‘%’ token
   25 |     printf("average = % f variance = % f\ n", average, variance);
      |                       ^
Pointers.c:25:43: error: stray ‘\’ in program
   25 |     printf("average = % f variance = % f\ n", average, variance);
      |                                         ^
Pointers.c:25:46: error: stray ‘\342’ in program
   25 |     printf("average = % f variance = % f\ n", average, variance);
      |                                            ^

Here's my original code

C

#include <stdio.h>
double average_variance(double data[], int numdata) 
{
	int i;
    double sqDiff = 0;
    double avg = averages(data, numdata);
    for (int i = 0; i<numdata; i++)
        sqDiff = (data[i] - numdata) * (data[i] - numdata);
    }
    return (sqDiff / numdata);

	double sum = 0;
	for (int i = 0; i<numdata; i++)
		sum += data[i];
	}
	return (sum / numdata);
}


int main(int argc, char ** argv)
{
    double values[] = {1,2,3,5,6,7};
    double variance;
    double avg = average_variance(values, 6, & variance);
    printf("average = % f variance = % f\ n", average, variance);

}

What I have tried:

I've tried to indent numerous times as It may be linked to a syntax issue but to no avail.

Posted 25-Feb-21 16:57pm

Member 15084336

Updated 25-Feb-21 20:24pm

CPallini

v2

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Patrice T · Answer 1 · 2021-02-25T18:01:00

C++

double average_variance(double data[], int numdata) 
{
	int i;
    double sqDiff = 0;
    double avg = averages(data, numdata);
    for (int i = 0; i<numdata; i++) // probably missing a { here
        sqDiff = (data[i] - numdata) * (data[i] - numdata);
    } // to match the ine here
    return (sqDiff / numdata);

	double sum = 0;
	for (int i = 0; i<numdata; i++) // probably missing a { here
		sum += data[i];
	} // to match the ine here
	return (sum / numdata);
}

Another problem, is the you can have only 1 return, bot two.

Advice: Learn to indent properly your code with professional editor, it show its structure and it helps reading and understanding. It also helps spotting structures mistakes.

C++

#include <stdio.h>
double average_variance(double data[], int numdata)
{
	int i;
	double sqDiff = 0;
	double avg = averages(data, numdata);
	for (int i = 0; i<numdata; i++)
		sqDiff = (data[i] - numdata) * (data[i] - numdata);
}
return (sqDiff / numdata); // position of this line indicate a problem with { and }

double sum = 0;
for (int i = 0; i<numdata; i++)
	sum += data[i];
}
return (sum / numdata);
}

int main(int argc, char ** argv)
{
	double values[] = {1,2,3,5,6,7};
	double variance;
	double avg = average_variance(values, 6, & variance);
	printf("average = % f variance = % f\ n", average, variance);

}

Indentation style - Wikipedia[^]

Professional programmer's editors have this feature and others ones such as parenthesis matching and syntax highlighting.
Notepad++ Home[^]
ultraedit[^]
Enabling Open Innovation & Collaboration | The Eclipse Foundation[^]

Rick York · Answer 2 · 2021-02-25T18:19:00

Patrice has given you some valid tips. I think you have several mistakes. You are missing the function definition for averages which was being called in average_variance. Also, average_variance needs to add the squares of the difference and the variance needs to be passed as an argument because that's how you are calling the function. Try this:

C++

double averages( double data[], int numdata )
{
	double sum = 0;
	for (int i = 0; i<numdata; i++)
    {
		sum += data[i];
	}
	return (sum / numdata);
}

double square( double val )
{
    return val * val;
}

double average_variance( double data[], int numdata, double & variance )
{
    double avg = averages( data, numdata );
    double sqDiff = 0;
    int i;
    for( i = 0; i < numdata; i++ )
    {
        sqDiff += square( data[i] - numdata );
    }
    variance = sqDiff / numdata;
    return avg;
}

#define ARG_COUNT 6

int main(int argc, char ** argv)
{
    double values[ARG_COUNT] = {1,2,3,5,6,7};
    double variance = 0;
    double avg = average_variance( values, ARG_COUNT, variance );
    printf( "average is %.3f and variance is %.3f\n", average, variance );
}

I almost forgot - the printf statement is wrong because the floating point format specifier is %f. If you want to display six digits after the decimal point you can use %.6f as the format specifier.

Help on determing issues with "expected identifier errors".

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