C Programming Tutorials : printf Format String

This post describes the use of Format String or Control String in printf function. An example of printf function is :
printf(" [FORMAT_STRING/CONTROL_STRING] ", arg1, arg2, ... );

Basic Structure of Format String or Control String in printf() Function:

% [FLAG] [WIDTH] . [PRECISION] [MODIFIER] [TYPE]

[FLAG]

- (minus)	Output is left-adjusted within the field. Remaining field will be blank. If no used, output is right-adjusted within the field. Works with: char, int , float, double, string.
+ (plus)	The sign of the numeric will be added before the numeric value i.e. if the numric value is positive, then a + sign will be added before the value. Works with: int, float, double Prints nothing with: char, string
0 (zero)	Causes leading zeros to appear. Works with: string, int, float, double. Prints nothing with:  char
# (with o or x)	Causes octal and hex items to be preceded by 0 (zero) and 0x , respectively. Works with: ( Octal ) %#o ( Hex ) %#x  or %#X
# (with e, f or g)	Causes decimal point to be present in all floating point numbers. Prevents the truncation of trailing zeros in g-type conversion.

Note: FLAGS can be mixed in any order

[WIDTH]

It is a +ve integer. It defines minimum field width.

Case 1: Length of data < [WIDTH]
        => Data is printed right-adjusted

Case 2: Length of data >= [WIDTH]
        => Entire data gets printed.

Note: If the [WIDTH] is given as -ve, then the - (minus) sign will be treated as flag

[PRECISION]

with String	Indicates no of charecters to be printed . If [PRECISION] > Length of data, entire String gets printed
with float, double and  long double	Indicates no. of digits to be printed after decimal point. If [PRECISION] < no. of digits after decimal point, rounded value will be printed. If [PRECISION] > no. of digits after decimal point, trailing zeros will be added.
with integer	If [PRECISION] > Length of data, prints (precision - length of data) no. of zeros.
with char	Does nothing

Note: [PRECISION] is a positive integer. If it is given as negative, then - (minus) sign + precision value + type gets printed.

[MODIFIERS]

h	for short integer
l	for long integers or double
L	for long double

[TYPE]

%a	Hexadecimal output in the form 0x h.hhhhhh p+ or - xxxx (C99 only)
%A	Hexadecimal output in the form 0x h.hhhhhh P+ or - xxxx (C99 only)
%x	Hexadecimal integer e.g. 0x5fc8
%X	Hexadecimal integer e.g. 0x5FC8
%o	Octal integer e.g. 0879
%c	type char e.g. 'a', 'A', '8' etc.
%d	decimal integer (signed ) e.g. 5, -4 etc.
%i	decimal integer (signed ) e.g. 6, -6 etc.
%u	unsigned decimal integer etc. 5, 6 etc.
%s	String e.g. "Do you like this article?" etc.
%e	scientific notation in the form d.xxxxxx e + or - xxx
%E	scientific notation in the form d.xxxxxx E + or - xxx
%f	float in the form d.xxxxxx e.g. 5.000000
%g	uses %e or %f whichever is shorter e.g. 100000, 1e+006, 0.0005, 5e-005 etc.
%G	uses %E or %f whichever is shorter
%p	Segment address (of pointer variable) in Hex. e.g. 12ao
%Fp	Segment address and offset address in Hex. e.g. 0001:00f2
%n	The associated argument must be a pointer to an integer. This specifier causes the number of charecters written (upto the point at which %n is entercounted) to be stored in that integer.
%%	prints a % sign

Note: [WIDTH] and [PRECISION] can be specified as a argument variable in printf() function. For example, printf( "%*.*s" , width, precision );

Few Examples of printf() Function

Assume, the following variable declarations:
char str[]="world";
int a = 50;
int b = -50;
float c = 4.5446;
unsigned int d = 5;

Note: The following examples are Tested on 32 bit GCC compiler

Statements	Output
printf("%s", str); printf("%-10s",str); printf("%2s",str); printf("%-2s",str); printf("%(-10)s",str); printf("%-(-10)s",str); printf("%20(-10)s",str); printf("%5s",str); printf("%-5s",str); printf("%+10s",str); printf("%010s",str);	world world world world (-10)s (-10)s (-10)s world world      world 00000world
printf("%-10d",a); printf("%+-10d",a); printf("%010d",a); printf("%0+10d",a); printf("%+0+10d",a); printf("%010d",b); printf("%#d",a); printf("%#x %x",a,a); printf("%#o %o",a,a); printf("%f",6); printf("%f",6.); printf("%f",6.0); printf("%f"); printf("%d");	50 +50 0000000050 +000000050 +000000050 -000000050 50 0x32 32 062 62 0.000000 6.000000 6.000000 0.000000 garbage value
printf("%10.2s",str); printf("%10.8s",str); printf("%10.-4s",str); printf("%1.-4s",str); printf("%10.5f",c); printf("%10.3f",c); printf("%10.9f",c); printf("%.5f",c); printf("%4.3d",a); printf("%.0d",a); printf("%d %d", 5, -5); printf("%i %i", 6, -6); printf("%u %u", 7, -7); printf("%5.3c",'d'); printf("%0c",'d'); printf("%I64"); printf("%s","Hello"); printf("%s","%d",5); printf("I64d",5); printf("%I64d",-5); printf("%d"+1,5);	wo      world -4s -4s 4.54460 4.545 4.54600000 4.54460 050 50 5 -5 6 -6 7 4294967289      d d prints nothing Hello %d 21474836485 25769803771 d

Confused, Have any questions?

Post your comments, questions, State your confusion to us.

We're here to help you.

Are you facing problems in writing C programs?

Hire us now!

Submit Your Problem Now

If You Enjoy This Post, Do us a favor: Share This Page

Subscribe via Email

Online C Programming Test

Test Description

Subject = C Programming
No of Questions = 20
Time = 30 mins
Total Marks = 20

Correct Answer = 1 mark
Wrong Answer = - 0.25 mark (negative)

Question 1: What is the output of the following C program?
int main(){
int i=5;
printf("%d %d %d %d %d",i++,i--,++i,--i,i);
}
(A) 45554
(B) 45545
(C) 56655
(D) Undefined results

Correct Answer is B.
Explanation:
The arguments in a function are pushed into the stack from left to right.
The evaluation is by popping out from the stack.
So, the evaluation is from right to left.
Hence, the order of execution is as follows: i, --i, ++i, i-- and finally i++

Question 2: What is the output of the following C program?
#define a 10
int main(){
#define a 50
int i=5;
printf("%d",a);
}
(A) 10
(B) 50
(C) Compiler Error
(D) Unexpected results

Correct Answer is B.
Explanation:
After executing #define a 10
main(){
#define 10 50
printf("%d",10); }

After executing next #define 10 50
main(){
printf("%d",50); }

Question 3: What is the output of the following C program?
int main(){
int i = -3, j = 2, k = 0, m;
m = ++i && ++j || ++k;
printf("%d %d %d %d",i,j,k,m);
}
(A) -2 3 0 1
(B) -3 3 0 1
(C) -2 3 1 1
(D) -3 3 1 1

Correct Answer is A.
Explanation:
Here the priority of && is greater than || so && will be executed first
m = (( ++i && ++j ) || ++k)
1st the value of i & j will be incremented by 1. But as because left of || becomes true, irrespective of the value of right side, the value of the total expression will be true. So value of k will not be incremented.

Question 4: What is the output of the following C program?
int main(){
int i = 4, j = 2;
int *p = &i, *q = &j;
int k = *p/*q;
printf("%d",k);
}
(A) 2
(B) 0
(C) Compiler Error
(D) Runtime Error

Correct Answer is C.
Explanation:
Compiler Error: Unexpected end of file in a comment
There should be space between / and * in line 4

Question 5: What error would the following function give on compilation?
f(int a, int b){
int a;
a = 20;
return a;
}
(A) Missing Parenthesis in return statement
(B) The function should be defined as int f(int a, intb)
(C) Redeclaration of a
(D) No Error

Correct Answer is C.
Explanation:
Compiler Error: variable a is declared twice.
One in formal arguments and another in body.
So, Complier Error: Redeclaration of a

Question 6: If n set the value 2, then the statement a[++n] = n++; will assign
(A) 3 to a[4]
(B) 2 to a[4]
(C) 3 to a[3]
(D) 4 to a[3]

Correct Answer is B.
Explanation:
a[++n] = n++;
Here post increment operator will return value of n 1st.
then n is incremented by 1
Now pre-increment operator increments the value of n by 1 first and then return it.
so here 2 will be assigned to a[4]

Question 7: What will be the output of the following C function?
printf("xy","ab","mn");
(A) xy
(B) xy ab mn
(C) garbage value
(D) Compiler error

Correct Answer is A.
Explanation:
It will only print "xy"
Other terms will not be printed.

Question 8: What will be the output of the following C program?
main(){
unsigned i = 3;
if(i>-1) printf("I like Qualitycoding");
else printf("I love QualityCoding");
}
(A) I like QualityCoding
(B) I love QualityCoding
(C) Compiler Error
(D) Unexpected Results

Correct Answer is B.
Explanation:
Comparison is to be done with both numbers as unsigned.
Unsigned representation of -1 is 32769

Question 9: What will be the output of the following C program?
main(){
printf(5+"Intelligent");
}
(A) Error
(B) Intelligent
(C) ligent
(D) Intell

Correct Answer is C.
Explanation:
printf function will print ligent.
first 5 charecters will not be printed.

Question 10: What will be the output of the following C program?
main(){
while(1) {
if( printf("%d", printf("%d") ) )
break;
else continue;
}
(A) Compilation Error
(B) Runtime Error
(C) Garbage Value
(D) prints nothing

Correct Answer is C.
Explanation:
As 'printf' is declared within a 'printf' it will print some garbage value. And as it is non zero value, so it will be executed and it will come out of the while loop.

Question 11: What will be the output of the following C program?
#define prod(a,b) a*b
main(){
int x=3, y=4;
printf("%d", prod(x+2, y-1) );
}
(A) 10
(B) 15
(C) 12
(D) Compilation Error

Correct Answer is A.
Explanation:
here prod(x+2,y-1) will be substituted by macro as x + 2 * y - 1 = 10

Question 12: What will be the output of the following C program?
main(){
printf("%d %d",sizeof(NULL), sizeof(" ") );
}
(A) 4 2
(B) 2 2
(C) 4 4
(D) 2 4

Correct Answer is A.
Explanation:
In stdio.h header file NULL is defined as 0. So sizeof(0) is 4.
But sizeof(" ") is 2. " " means two charecters space i.e. ' ' and null charecter '\0'

Question 13: What will be the output of the following C program?
main(){
char str1[]="Hello";
char str2[]="Hello";
if(str1==str2)
printf("equal");
else
printf("unequal");
}
(A) equal
(B) unequal
(C) Compilation Error
(D) Runtime Error

Correct Answer is B.
Explanation:
As str1 and str2 are comparing their base addresses and base addresses cann't be equal. So it will print 'unequal'.

Question 14: What will be the output of the following C program?
main(){
printf("%c",3["beautiful"]);
}
(A) 3rd charecter from left i.e. prints a
(B) 3rd charecter from right i.e. prints f
(C) 4th charecter from left i.e. prints u
(D) 4th charecter from right i.e. prints i

Correct Answer is C.
Explanation:
prints 4th charecter from left i.e. prints u.
3["beautiful"] is equivalent to "beautiful"[3]

Question 15: What will be the output of the following C program?
main(){
float a=2.2, c=2.5;
if(a==2.2) printf("Hello,");
if(c==2.5) printf("How are you");
}
(A) Hello, How are you
(B) Hello,
(C) How are you
(D) prints nothing

Correct Answer is C.
Explanation:
comparing floating point value with == operator always gives false value except when fractional part is either .5 or .0

Question 16: What will be the output of the following C program?
main(){
if(0,1,2,3)
printf("I like QualityCoding");
else
printf("I love QualityCoding");
}
(A) I like QualityCoding
(B) I love QualityCoding
(C) Compilation Error
(D) Runtime Error

Correct Answer is A.
Explanation:
The condition if(0,1,2,3) will evaluated to if(3) which is true because 3!=0 is true.
Comma operator will return 3 here i.e. the right most value.

Question 17: What will be the output of the following C program?
main(){
printf("%s",0);
}
(A) (false)
(B) 0
(C) (null)
(D) Compilation Error

Correct Answer is C.
Explanation:
It will print (null).

Question 18: What will be the output of the following C program?
main(){
int i, a[]={000,001,010,011,100,101,110,111};
for(i=0;i<8;i++)
printf("%d ",a[i]);
}
(A) 0 1 10 11 100 101 110 111
(B) 0 1 8 9 64 65 72 73
(C) 0 1 8 9 64 65 110 111
(D) 0 1 8 9 100 101 110 111

Correct Answer is D.
Explanation:
The value of 010 and 011 will be 8 and 9. because these two numbers 010 and 011 are in octal i.e. preceeded by 0.
So, after storing it into the array, it will be converted into decimal value.

Question 19: What will be the output of the following C program?
main(){
int i=5;
printf("%d ",i++++++i);
}
(A) 10
(B) 11
(C) Unexpected result
(D) Compiler Error

Correct Answer is D.
Explanation:
The expression is parsed as i++ ++ + i which is illegal. So, compiler error.

Question 20: What will be the output of the following C program?
main(){
int i=10,*p=&i;
int c=i/*p; /*Result of division operation is stored in variable c*/;
printf("%d",c);
}
(A) 1
(B) 10
(C) unexpected results
(D) Compiler Error

Correct Answer is B.
Explanation:
line 3 will get evaluated as int c=i;
comment line in line 3 is /*p; /*Result of division operation is stored in variable c*/
After this comment line there is a semi-colon(;). Hence, answer is 10

C Program to find Armstrong Number

Definition of Armstrong Number:
A number is Armstrong if the sum of cubes of individual digits of a number is equal to the number itself.

Example:
0, 1, 153, 370, 371, 407 etc.

Explanation:
Let us take 153.

So, the sum of cubes of individual digits

= 1^{3 +} 5^{3 +} 3³

= 1 + 125 + 27

= 153

Programming Approach (Algorithm):

Step 1: number = <Entered Number>

Step 2: Right Most Digit = number % 10

Step 3: number = number / 10

Step 4: Sum = Sum + Cube of Right Most Digit

Step 5: Repeat Step 2, 3 & 4 until number = 0

Step 6: if Sum = <Entered Number> then return 1

                                 // <Entered Number> is Armstrong number

        else return 0 // <Entered Number> is not an Armstrong number

C Program:

/*
 * PROGRAM NAME: Armstrong Number
 * SOURCE: http://CProgrammingOnline.blogspot.com
 */

/* HEADER FILE DECLARATION SECTION */
#include<stdio.h>

/* FUNCTION PROTOTYPE DECLARATION SECTION */
_Bool armstrong(int );

/* MAIN PROGRAM SECTION */
    int main(char* argv[], int argc){
    int number;    //Used to store the given number

    printf("\nProgram to Test Whether a Number is Armstrong Number or Not\n");
    printf("Enter an Integer Value (Max Size = %d bytes): ",sizeof(int));
    scanf("%d",&number);

    if(armstrong(number)==1){
        printf("\n%d is an armstrong number.",number);
    } else {
        printf("\n%d is not an armstrong number.",number);
    }
}

/*
 * FUNCTION SIGNATURE: _Bool armstrong(int );
 * INPUT: int number
 * OUTPUT: 1 if number(INPUT) is an armstrong number
 *         0 if number(INPUT) is not an armstrong number
 */
_Bool armstrong(int number){
    int stored_number = number;  //Used to Store the given number
    int sum = 0;                       //Used to Store the sum of digits
    char digit;                       //Used to Store the digits

    while(stored_number > 0){
        digit = stored_number % 10;
        sum += digit * digit * digit;
        stored_number /= 10;
    }

    if(sum == number){
        return 1;
    } else {
        return 0;
    }
}

Confused, Have any questions?
Post your comments, questions. State your confusion to us. We're here to help you.

Are you facing problems in writing C programs?
Hire us now!
Submit Your Problem Now

Subscribe via Email

Copy Text File using C Program

#include<stdio.h>

int main(int argc, char* argv[]){
    FILE *sfp;                   // Source File Pointer
    FILE *tfp;                   // Target File Pointer
    char SourceFilename[255];    // Stores the source file name
    char TargetFilename[255];    // stores the target file name
    char ch;                     // used to read charecter
                                 //       from source file name

    /* Get sourcefile name and targetfile name from the user */
    printf("\nEnter Soucefile Name: ");
    gets(SourceFilename);
    fflush(stdin);
    printf("Enter Targetfile Name (NOTE : If Targetfile already exists, its content will be overwritten): ");
    gets(TargetFilename);

    /* Open source file in read mode and target file in write mode */
    sfp=fopen(SourceFilename,"r");
    if(sfp==NULL){
        printf("\nERROR: Failed to open %s in read mode. Closing the program...",SourceFilename);
        return;
    }
    tfp=fopen(TargetFilename,"w");
    if(tfp==NULL){
        printf("\nERROR: Failed to open %s in write mode. Closing the program...",TargetFilename);
        return;
    }

    /* read charecter by charecter from sourcefile
       and write those charecters accordingly in targetfile */
    while(1){
        ch=fgetc(sfp);
        if(ch==EOF){
            break;
        }else{
            fputc(ch,tfp);
        }
    }

    /* close file pointers */
    fclose(sfp);
    fclose(tfp);
}