Function pointers

Function pointers is the focus of this chapter in the C learning path. Before writing a large program, practice this one idea with a small input and a visible result. This chapter connects the idea to the code patterns used in C applications.

A loop repeats a block of instructions over a collection. An if condition chooses whether a statement runs. Modulus returns a remainder. Checking remainder zero tests for an even number.

• Function pointers
• Key vocabulary and inputs
• Reading the output step by step
• Predicting edge cases
• Applying this skill in a small project

The important learning habit is to connect a technical word with an observable behavior, not simply memorize its definition. Make one prediction before pressing Run.

Follow one value at a time: check its condition, possibly print it, then move to the next.

1. Line 1 #include <stdio.h> Read this instruction before running the next line. Watch what it assigns, checks or prints.
2. Line 2 int main(void){ Read this instruction before running the next line. Watch what it assigns, checks or prints.
3. Line 3 int values[]={9,12,15}; Read this instruction before running the next line. Watch what it assigns, checks or prints.
4. Line 4 for(int i=0;i<3;i++){if(values[i]%2==0)printf("%d is even\n",values[i]);} Read this instruction before running the next line. Watch what it assigns, checks or prints.
5. Line 5 printf("Count: 3\n"); Read this instruction before running the next line. Watch what it assigns, checks or prints.
6. Line 6 return 0; Read this instruction before running the next line. Watch what it assigns, checks or prints.
7. Line 7 } Read this instruction before running the next line. Watch what it assigns, checks or prints.

Try it: Change the numeric input in line one or two. Say aloud which output line must change, and why, before you rerun.

Example 1: Function pointers — experiment 1

#include <stdio.h>
int main(void){
int values[]={9,12,15};
for(int i=0;i<3;i++){if(values[i]%2==0)printf("%d is even\n",values[i]);}
printf("Count: 3\n");
return 0;
}

12 is even
Count: 3

Why this output? A loop repeats a block of instructions over a collection. An if condition chooses whether a statement runs. Follow one value at a time: check its condition, possibly print it, then move to the next. Change one input value, rerun the program and compare the result before moving on.

Example 2: Function pointers — experiment 2

#include <stdio.h>
int main(void){
int values[]={10,13,16};
for(int i=0;i<3;i++){if(values[i]%2==0)printf("%d is even\n",values[i]);}
printf("Count: 3\n");
return 0;
}

10 is even
16 is even
Count: 3

Why this output? A loop repeats a block of instructions over a collection. An if condition chooses whether a statement runs. Follow one value at a time: check its condition, possibly print it, then move to the next. Change one input value, rerun the program and compare the result before moving on.

Example 3: Function pointers — experiment 3

#include <stdio.h>
int main(void){
int values[]={11,14,17};
for(int i=0;i<3;i++){if(values[i]%2==0)printf("%d is even\n",values[i]);}
printf("Count: 3\n");
return 0;
}

14 is even
Count: 3

Why this output? A loop repeats a block of instructions over a collection. An if condition chooses whether a statement runs. Follow one value at a time: check its condition, possibly print it, then move to the next. Change one input value, rerun the program and compare the result before moving on.

1. Predict: What is printed when you run the first example without changing anything? Compare every character against the output panel.
2. Modify: Change the input number. Does the program compute or display a different value? Explain why.
3. Explain: Which statement produces the visible output, and which statement stores or transforms the input?
4. Debug: Intentionally remove a bracket or quote. Read the error and restore the missing character.

Worked answer

The final example is a complete, runnable model solution for this level. Read the input, execute the operation and check the output: no line should appear by accident.

#include <stdio.h>
int main(void){
int values[]={11,14,17};
for(int i=0;i<3;i++){if(values[i]%2==0)printf("%d is even\n",values[i]);}
printf("Count: 3\n");
return 0;
}

14 is even
Count: 3

Mini-project: add the core idea from Function pointers to a small C learning application. Name the input, validate it, transform it, and print a result. Keep the behavior simple enough to test by hand.

For framework-specific production examples and prerequisite setup, see the course overview and bundled starter project. The locally run examples above demonstrate underlying programming concepts, not every advanced feature named by this chapter.