Write a program that declares three one-dimensional arrays named current, resist
ID: 3933503 • Letter: W
Question
Write a program that declares three one-dimensional arrays named current, resistance, and voltage.
Each array should be declared in main() and be capable of holding 10 double-precision numbers.
The numbers to be stored in current are 10.62, 14.89, 13.21, 16.55, 18.62, 9.47, 6.58, 18.32, 12.15, and 3.98.
The numbers to be stored in resistance are 4, 8.5, 6, 7.35, 9, 15.3, 3, 5.4, 2.9, and 4.8. Have your program pass
these three arrays to a function called calcVolts(), which calculates the elements in the voltage array as
the product of the equivalent elements in the current and resistance arrays: for example, voltage[1] = current[1] * resistance[1].
After calcVolts() has put values in the voltage array, display the values in the array from within main(). Write the calcVolts() function by using pointers.
Explanation / Answer
#include <stdio.h>
void calcvolts(double *current,double *resistance,double *voltage)// to compute voltage values
{
int i;
for(i=0;i<10;i++)
*(voltage+i)=*(current+i)+*(resistance+i);
}
int main()
{
double current[]={10.62, 14.89, 13.21, 16.55, 18.62, 9.47, 6.58, 18.32, 12.15,3.98};
double resistance[]={4, 8.5, 6, 7.35, 9, 15.3, 3, 5.4, 2.9, 4.8},voltage[10];
int i;
calcvolts(current,resistance,voltage);//function call passing the arrays
printf("content of voltage array is ");
for(i=0;i<10;i++)
printf(" voltage[%d]=%lf ",i,voltage[i]);
return 0;
}
output:
voltage[9]=4621132318645797519
sh-4.3$ gcc -o main *.c
sh-4.3$ main
content of voltage array is
voltage[0]=14.620000
voltage[1]=23.390000
voltage[2]=19.210000
voltage[3]=23.900000
voltage[4]=27.620000
voltage[5]=24.770000
voltage[6]=9.580000
voltage[7]=23.720000
voltage[8]=15.050000
voltage[9]=8.780000
sh-4.3$
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