printf

printf is a formated print function, included in stdio (standard input output) library. To use this, you need to #include <stdio.h>.

• Example:

  #include <stdio.h>
  
  int main(void) {
    int          varInt = -100;
    unsigned int varUInt = 44;
    double       varDouble = 1.23451234512345;
  
    printf("Just character string!\n");
    printf("integers: %d  %u\n", varInt, varUInt);
    printf("Float: %f %3f %.4f %.15f\n", varDouble, varDouble, varDouble, varDouble);
  
    return(0);
  }
  

  • printf("character string")
  • printf("character string with format specifier", nameOfVar1, nameOfVar2,...)
    
    • Format specifier starts from %.
    • Format specifiers are the place holder, and they are substitued by the values of variables.
    • Number of % in the character string shuld match with the number of variables.
  • Try to use a wrong format specifier. Change %d in the 2nd printf to %u, and check what happens.
• Common format specifiers

  specifier	meaning	var type
  %d	Signed integer	short, int
  %ld	Signed long integer	long
  %lld	Signed long long integer	long long
  %u	Unsigned integer	unsigned short, unsigned int
  %lu	Unsigned long integer	unsigned long
  %llu	Unsigned long long integer	unsigned long long
  %f	Decimal floating-point	float, double

• Fancier formatting

  An integer inserted right after % specifies field width.
  An integer behind . means precision

    int    i = 123;
    doubel x = 0.123456789;
  

  Format	Printed string	comments
  printf("%d",i);	"123"
  printf("%5d",i);	" 123"	right adjusted, field width 5
  printf{"%8.5f",x);	" 0.12346"	field width 8, precision 5

• Special characters

  \t	tab
  \n	newline

• Exercise:
  Make a program to check following:
  • What happens if you assign 1.732 to a integer variable?
  • What happens if you assign a negative number to an unsigned integer? Try with -1 and -2.