C Exercises: Implement the printf() function

Last update on August 07 2025 12:52:35 (UTC/GMT +8 hours)

7. Variadic printf() Implementation Challenges

Write a C program to implement the printf() function using variadic functions.

Sample Solution:

C Code:

#include <stdio.h>

#include <stdarg.h>

void printValues(char * format, ...) {
  va_list argList;
  va_start(argList, format);

  printf("The values are: ");
  while ( * format != '\0') {
    if ( * format == 'd') {
      int val = va_arg(argList, int);
      printf("%d ", val);
    }
    format++;
  }
  printf("\n");
  va_end(argList);
}
int main() {
  int x = 10;
  int y = 20;
  int z = 30;
  printValues("ddd", x, y, z);
  return 0;
}

Sample Output:

The values are: 10 20 30

Explanation:

This above program defines the function printValues () that takes a string format and a variable number of arguments. Inside the function, we use the va_list, va_start(), va_arg(), and va_end() macros from the stdarg.h library to iterate over the variable arguments.

We then loop through each character in the format string and check if it is a % character. If it is, we use va_arg() to get the next argument as an integer and print it with printf(). If it is not a % character, we print it using printf().

Flowchart:

For more Practice: Solve these Related Problems:

Write a C program to implement a simplified version of printf() that supports %d and %s format specifiers using variadic functions.
Write a C program to create a custom printf() that handles %f conversions and prints floating-point numbers accurately.
Write a C program to implement a variadic function that mimics printf() by printing multiple integers separated by spaces.
Write a C program to develop a minimal printf()-like function using variadic arguments that supports a limited set of format specifiers with error checking.

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