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Let us look at some ways to use variables.

The problem: Read in a header value and then up to three triples of numbers representing the three dimensional point (x, y, z) and compute and print the distance from the origin of each point (in reverse order, except for the separate variable approach).

• Separate input, processing, and output. (Assume that you might have to process and output the items in reverse order).
• Avoid statement and control structure redundancy.

The following approaches are used.

• Individual double variables
• Parallel arrays of double elements
• Arrays of record structures of double named fields

One way is to use individual variables.
double x0,y0,z0,d0; // not needed double x1,y1,z1,d1; double x2,y2,z2,d2; double x3,y3,z3,d3;
Note: This way does not scale up well and requires iteration without loops.

Another way uses parallel arrays. Here the list starts at position 1 so MAX1+1 parallel array elements are needed.
const int MAX1 = 3; double x[MAX1+1],y[MAX1+1],z[MAX1+1],d[MAX1+1];
An array is needed for each of the four variables for x1, y1, z1, d1 so that x2, y2, z2, d2, etc. are not needed.

Parallel arrays are needed but it would be nice to combine each element of each array at the same position into a record structure.

Here is a way using structures.
const int MAX1 = 3; struct Point { double x, y, z, d; }; struct Point p[MAX1+1];
The record structure is given the name Point and an array of Point elements (each with an x, y, z, and d) is declared.

Here the list starts at position 1 so MAX1+1 array elements of record structures are needed.

Let us compare the three methods for double variables, array of double elements, arrays of record structures.

• x1 is x[1] is p[1].x
• y1 is y[1] is p[1].y
• z1 is z[1] is p[1].z
• d1 is d[1] is p[1].d
• x2 is x[2] is p[2].x
• y2 is y[2] is p[2].y
• ... and so on ...

In the case of parallel arrays of elements and arrays of records and index variable/expression can be used to abstract and generalize.

Here are some images to depict the three ways.

• Individual variables
• Parallel array of variables
• Array of record variables

The question mark "?" in each variable memory location is used to show that these variables are not yet initialized with a value.

Note: The individual variables x0, y0, z0, and d0 are not needed, but since the other methods start the list at index 1 and not at 0, these individual variables are shown for visual comparison.

Four parallel arrays of variables are needed: x, y, z and d.

One array of record variables is needed: p. Each element of array p has named fields (variables) x, y, z and d.

All of the ways start by getting the header value.
scanf("%d",&len1); printf("Input %d values (header loop).\n",len1);

The individual variable method requires iteration without a loop. See the code below to see how this can be done.

Here is a comparison of the input loop for arrays and records.
for (int pos1=1; pos1 <= len1; pos1++) { scanf("%lf %lf %lf",&x[pos1],&y[pos1],&z[pos1]); }

for (int pos1=1; pos1 <= len1; pos1++) { scanf("%lf %lf %lf",&p[pos1].x,&p[pos1].y,&p[pos1].z); }

Here is a comparison of the process loop for arrays and records.
for (int pos1=1; pos1 <= len1; pos1++) { d[pos1] = sqrt(pow(x[pos1],2)+pow(y[pos1],2)+pow(z[pos1],2)); }

for (int pos1=1; pos1 <= len1; pos1++) { p[pos1].d = sqrt(pow(p[pos1].x,2)+pow(p[pos1].y,2)+pow(p[pos1].z,2)); }

Here is a comparison of the output loop for arrays and records.
for (int pos1=len1; pos1 >= 1; pos1--) { printf("%d. Length to (%0.2lf,%0.2lf,%0.2lf) is %0.2lf\n",pos1,x[pos1],y[pos1],z[pos1],d[pos1]); }

for (int pos1=len1; pos1 >= 1; pos1--) { printf("%d. Length to (%0.2lf,%0.2lf,%0.2lf) is %0.2lf\n",pos1,p[pos1].x,p[pos1].y,p[pos1].z,p[pos1].d); }

Note: The individual variables x0, y0, z0, and d0 are not needed. See notes above. Here is the C code.
#include <stdio.h> #include <math.h> int main() { int pos1,len1; double x0,y0,z0,d0; // not needed double x1,y1,z1,d1; double x2,y2,z2,d2; double x3,y3,z3,d3; // input scanf("%d",&len1); printf("Input %d values (header loop).\n",len1); pos1 = 0; if (pos1 != len1) { pos1 += 1; scanf("%lf %lf %lf",&x1,&y1,&z1); if (pos1 != len1) { pos1 += 1; scanf("%lf %lf %lf",&x2,&y2,&z2); if (pos1 != len1) { pos1 += 1; scanf("%lf %lf %lf",&x3,&y3,&z3); } } } // process printf("Process %d values.\n",len1); pos1 = 0; if (pos1 != len1) { pos1 += 1; d1 = sqrt(pow(x1,2)+pow(y1,2)+pow(z1,2)); if (pos1 != len1) { pos1 += 1; d2 = sqrt(pow(x2,2)+pow(y2,2)+pow(z2,2)); if (pos1 != len1) { pos1 += 1; d3 = sqrt(pow(x3,2)+pow(y3,2)+pow(z3,2)); } } } // output printf("Output %d values.\n",len1); pos1 = 0; if (pos1 != len1) { pos1 += 1; printf("%d. Length to (%0.2lf,%0.2lf,%0.2lf) is %0.2lf\n",pos1,x1,y1,z1,d1); if (pos1 != len1) { pos1 += 1; printf("%d. Length to (%0.2lf,%0.2lf,%0.2lf) is %0.2lf\n",pos1,x2,y2,z2,d2); if (pos1 != len1) { pos1 += 1; printf("%d. Length to (%0.2lf,%0.2lf,%0.2lf) is %0.2lf\n",pos1,x3,y3,z3,d3); } } } return 0; }

Here are some examples of input and output for the above program code.

Here is an example input.
3 1.1 2.2 3.3 2.2 3.3 4.4 3.3 4.4 5.5
For the above example input, here is the expected output.
Input 3 values (header loop). Process 3 values. Output 3 values. 1. Length to (1.10,2.20,3.30) is 4.12 2. Length to (2.20,3.30,4.40) is 5.92 3. Length to (3.30,4.40,5.50) is 7.78
Here is an example input.
2 1.2 3.4 5.6 2.3 4.5 6.7
For the above example input, here is the expected output.
Input 2 values (header loop). Process 2 values. Output 2 values. 1. Length to (1.20,3.40,5.60) is 6.66 2. Length to (2.30,4.50,6.70) is 8.39
Here is an example input.
1 0.1 2.3 4.5
For the above example input, here is the expected output.
Input 1 values (header loop). Process 1 values. Output 1 values. 1. Length to (0.10,2.30,4.50) is 5.05
Here is an example input.
0
For the above example input, here is the expected output.
Input 0 values (header loop). Process 0 values. Output 0 values.

Four parallel arrays of variables are needed: x, y, z and d. Here is the C code.
#include <stdio.h> #include <math.h> int main() { int pos1,len1; const int MAX1 = 3; double x[MAX1+1],y[MAX1+1],z[MAX1+1],d[MAX1+1]; // input scanf("%d",&len1); printf("Input %d values (header loop).\n",len1); for (int pos1=1; pos1 <= len1; pos1++) { scanf("%lf %lf %lf",&x[pos1],&y[pos1],&z[pos1]); } // process printf("Process %d values.\n",len1); for (int pos1=1; pos1 <= len1; pos1++) { d[pos1] = sqrt(pow(x[pos1],2)+pow(y[pos1],2)+pow(z[pos1],2)); } // output for (int pos1=len1; pos1 >= 1; pos1--) { printf("%d. Length to (%0.2lf,%0.2lf,%0.2lf) is %0.2lf\n",pos1,x[pos1],y[pos1],z[pos1],d[pos1]); } return 0; }

Here are some examples of input and output for the above program code.

Here is an example input.
3 1.1 2.2 3.3 2.2 3.3 4.4 3.3 4.4 5.5
For the above example input, here is the expected output.
Input 3 values (header loop). Process 3 values. 3. Length to (3.30,4.40,5.50) is 7.78 2. Length to (2.20,3.30,4.40) is 5.92 1. Length to (1.10,2.20,3.30) is 4.12
Here is an example input.
2 1.2 3.4 5.6 2.3 4.5 6.7
For the above example input, here is the expected output.
Input 2 values (header loop). Process 2 values. 2. Length to (2.30,4.50,6.70) is 8.39 1. Length to (1.20,3.40,5.60) is 6.66
Here is an example input.
1 0.1 2.3 4.5
For the above example input, here is the expected output.
Input 1 values (header loop). Process 1 values. 1. Length to (0.10,2.30,4.50) is 5.05
Here is an example input.
0
For the above example input, here is the expected output.
Input 0 values (header loop). Process 0 values.

One array of record variables is needed: p. Each element of array p has named fields (variables) x, y, z and d. Here is the C code.
#include <stdio.h> #include <math.h> int main() { int pos1,len1; const int MAX1 = 3; struct Point { double x, y, z, d; }; struct Point p[MAX1+1]; // input scanf("%d",&len1); printf("Input %d values (header loop).\n",len1); for (int pos1=1; pos1 <= len1; pos1++) { scanf("%lf %lf %lf",&p[pos1].x,&p[pos1].y,&p[pos1].z); } // process printf("Process %d values.\n",len1); for (int pos1=1; pos1 <= len1; pos1++) { p[pos1].d = sqrt(pow(p[pos1].x,2)+pow(p[pos1].y,2)+pow(p[pos1].z,2)); } // output for (int pos1=len1; pos1 >= 1; pos1--) { printf("%d. Length to (%0.2lf,%0.2lf,%0.2lf) is %0.2lf\n",pos1,p[pos1].x,p[pos1].y,p[pos1].z,p[pos1].d); } return 0; }

Here are some examples of input and output for the above program code.

Here is an example input.
3 1.1 2.2 3.3 2.2 3.3 4.4 3.3 4.4 5.5
For the above example input, here is the expected output.
Input 3 values (header loop). Process 3 values. 3. Length to (3.30,4.40,5.50) is 7.78 2. Length to (2.20,3.30,4.40) is 5.92 1. Length to (1.10,2.20,3.30) is 4.12
Here is an example input.
2 1.2 3.4 5.6 2.3 4.5 6.7
For the above example input, here is the expected output.
Input 2 values (header loop). Process 2 values. 2. Length to (2.30,4.50,6.70) is 8.39 1. Length to (1.20,3.40,5.60) is 6.66
Here is an example input.
1 0.1 2.3 4.5
For the above example input, here is the expected output.
Input 1 values (header loop). Process 1 values. 1. Length to (0.10,2.30,4.50) is 5.05
Here is an example input.
0
For the above example input, here is the expected output.
Input 0 values (header loop). Process 0 values.

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