While Loop Structure while (condition) { …. // This line will process when while condition is true }

While Loop

Structure

while (condition){

…. // This line will process when while condition is true

}

do - While Loop

Structure

do{

…. // This line will process first then check while condition

} while (condition); // if while condition is true then process inside do again

For Loop

Structure

for (initial; condition; increment){

…. // This line will process when for condition is true

}

initial = initialize value of index (do at once)Condition = check whether true or falseIncrement = after process then define the increment or decrement of index

Nested For Loop

Structure

for (initial; condition; increment){

…. (outer loop line)for (initial; condition; increment){

…. (inner loop line)}….. (outer loop line)

}

How to use “FOR” instead of “WHILE” in the following example

main(){int i;i =1;while(i <= 10) {

printf(“%d”,i);i = i+1;

}}

How to use “FOR” instead of “WHILE” in the following example

main(){int i;i =1;while(i <= 10) {

printf(“%d”,i);i =i+1;

}}

main(){int i;

for (i=1;i <= 10;i=i+1) {

printf(“%d”,i); }

}

A1 : Given The Following Code


i = 0 ; max = 6 ;

Solution:Loop1 : i=0, max=6 0Loop2 : i=3, max=6 3Loop3 : i=6, max=6


i = 0 ; max = 6 ;

Result: 03


i = -3 ; max = 6 ;

Solution:Loop1 : i=-3, max=6 -3Loop2 : i=0, max=6 0Loop3 : i=3, max=6 3Loop4 : i=6, max=6


i = -3 ; max = 6 ;

Result: -303


i = 2 ; max = ++i +i +3 ;


++i + i +3 = 3+3+3 9


i = 2 ; max = ++i +i +3 ;

Result: 36


max = 8 ; i = max/2;



max = 8 ; i = max/2;

Result: 47

A2 : What is the output of the following program?


Result: x is 3 (n=7)

Solution:Loop1 : x=15, n=4 13Loop2 : x=13, n=4 11Loop3 : x=11, n=4 9Loop4 : x=9, n=4 7Loop5 : x=7, n=4 5Loop6 : x=5, n=4 3Loop7 : x=3, n=4

B9 : What is the output of the following program?

B9 : What is the output of the following program?#include<stdio.h> int main(void) { int x; for(x=3;x<20;x++) { if(x==5) continue; if(x==10) break; printf("%d ",x); } return 0; }

Result: b. 3 4 6 7 8 9

Output Loop1(x=3) 3 Loop2(x=4) 4 Loop3(x=5) continue; Loop4(x=6) 6 Loop5(x=7) 7 Loop6(x=8) 8 Loop7(x=9) 9 Loop3(x=10) Break;

B11 : In the following C program, how many times will i be printed out on the screen?

B11 : In the following C program, how many times will i be printed out on the screen?

#include<stdio.h> int main(void) { int i=10; do{ printf("i"); i++; }while(i<10); }

Result: d. 1

Output Loop1(i=10) i Loop2(i=11)

ArrayStructure

int digits[10] = {1,2,3,4,5,6,7,8,9,10}; static float x[6] = { 0,0.25,0,0.50,0,0}; char color[] = {'R','E','D'}; char color2[] = "RED"

digits[0] = 1 x[0] = 0 color[0] = 'R‘digits[1] = 2 x[1] = 0.25 color[1] = 'E' digits[2] = 3 x[2] = 0 color[2] = 'D' digits[3] = 4 x[3] = 0.50 digits[4] = 5 x[4] = 0 color2[0] = 'R' digits[5] = 6 x[5] = 0 color2[1] = 'E' digits[6] = 7 color2[2] = 'D' digits[7] = 8 color2[3] = '\0' digits[8] = 9 digits[9] = 10

Link to Bubble Sort Example

B5:

B5:

Solution:

a. int a[3]={2,3,4};

b. int a[] = {2,3,4};

c. int a[5] = {2,3,4};

d. int a[3] = {2,3,4,0,0};

2 3 4

2 3 4 0 0

2 3 4

2 3 4 0 0

2 3 4

B6:

5 Rows

4 Columns

B6:

Solution:

a. int array[10][4]; array with 10 rows and 4 columns

b. int array[4][10]; array with 4 rows and 10 columns

c. int array[4,10]; ERROR

d. int a {10} {4}; ERROR

B18:

B10:

B10:

String• is defined as an ARRAY of CHARACTERS• String is terminated by a special character

which is null parameter (\0).• We declare a string by

char color[] = “blue”;char color[] = {‘b’, ‘l’, ‘u’, ‘e’,‘\0’}; char msg[10] = “Computer”;char msg[10] = {‘C’,’o’,’m’,’p’,’u’,’t’,’e’,’r’,’\0’};

String Input Functions • scanf (“%s”,name); Scan until first space

• gets(name); Scan until end of line

• sscanf(data,”%s”,name); Read from data and keep string in name

EXAMPLE: This is a Bookscanf : Thisgets: This is a Booksscanf : This

String Output Functions #include <stdio.h>main(){

int i;char name[]= “Smith”;printf(“Name = %s”, name);puts(name);for(i=0;i<5;i++){

printf(“%c”, name[i]);}

}

Link to Example of ctype.h

string.hstrcat(); Appends the string pointed to by str2 to the end of the string pointed to by str1. strncat(); Appends the string pointed to by str2 to the end of the string pointed to by str1 up to n characters long.strchr(); Searches for the first occurrence of the character c (an unsigned char) in the string pointed to by the argument str. strcmp(); Compares the string pointed to by str1 to the string pointed to by str2. strncmp(); Compares at most the first n bytes of str1 and str2. Stops comparing after the null character. strcpy(); Copies the string pointed to by str2 to str1. strncpy(); Copies up to n characters from the string pointed to by str2 to str1.

string.hstrcspn(); Finds the first sequence of characters in the string str1 that does not contain any character specified in str2. strlen(); Computes the length of the string str up to but not including the terminating null character. strpbrk(); Finds the first character in the string str1 that matches any character specified in str2. strrchr(); Searches for the last occurrence of the character c (an unsigned char) in the string pointed to by the argument str. strspn(); Finds the first sequence of characters in the string str1 that contains any character specified in str2. strstr(); Finds the first occurrence of the entire string str2 (not including the terminating null character) which appears in the string str1. strtok(); Breaks string str1 into a series of tokens. strxfrm(); Transforms the string str2 and places the result into str1.

B13:

B13:

B21:

B21:

Solution

printf("%d %c %c\n", toupper('a')+1, toupper('b')+1, tolower('C')-2);

toupper('a')+1 A (65) +1 = 66 %d 66

toupper('b')+1 B(66) +1 = 67 %c C

tolower('C')-2); c(99) – 2 = 97 %c a

B14:

B14:

B22:

B22:

Solution

a. string.h This header file defines several functions to manipulate C strings and arraysb. stdio.h This header file for the standard functions that deal with input and outputc. stdlib.h This header defines several general purpose functions, including dynamic memory management and randomd. ctype.h This header allow the programmer to check, compare, and manipulate characters

A5:

A5: Write a single statement to accomplish each of the following. Assume that variables c (which stores a character) is of type int, and arrays s1[100] and s2[100] are of type char.

5.1 Convert the character stored in variable c to an uppercase letter. Assign the result to variable c.


5.1 Convert the character stored in variable c to an uppercase letter. Assign the result to variable c.

#include <ctype.h>main() { int c = 'a'; c = toupper(c); printf("%c",c);}

Answer


5.2 Convert the string “8.635” to double and print the value.


5.2 Convert the string “8.635” to double and print the value.

#include <stdlib.h>main() { char s2[100] = "8.635"; printf("%.3f",atof(s2));

}

Answer


5.3 Copy the string stored in array s2 into array s1.


5.3 Copy the string stored in array s2 into array s1.

#include <string.h>main() { char s1[100], s2[100] = "8.635"; strcpy (s1,s2);

printf("%s",s1);}

Answer


5.4 Determine the length of the string in s1. Print the result.


5.4 Determine the length of the string in s1. Print the result.

#include <string.h>main() { char s1[100] = "8.635"; printf("%d",strlen(s1));

}

Answer

Why?

• Main reason– Often be used more than once in a program– Fit naturally with a top-down design approach– Provide a natural method for dividing a

programming task among team– Can be tested individually

Example: No Output

void print_hi(){

printf(“Hi”);

}

No Output

Example: No Input

void print_hi(){

printf(“Hi”);

}

void print_hi(void){

printf(“Hi”);

}

No Input

Example: Input as Integer

int factorial(int n){

int ans=1;int i;for (i=2;i<n;i++){

ans*=i;}return ans;

}

n is input

Example: Output as Integer

int factorial(int n){

int ans=1;int i;for (i=2;i<n;i++){

ans*=i;}return ans;

}

Output as Integer

Output value of ans

Example: Two Input

void print_chars(char ch, int n){

int i;for (i=0;i<n;i++){

printf(“%c”,ch);}

}

Function Prototypes

double cal_area(double r);void main(){

…area= cal_area(r);…

}

double cal_area(double r){

return 3.14*r*r;}


return 3.14*r*r;}

void main(){

…area= cal_area(r);…

}

Global vs Local Variable

int pi=3.14;double cal_area(double r){

return pi*r*r;}void main(){

…area= cal_area(r);cir= 2*pi*r;

}


int pi=3.14;return pi*r*r;

}

void main(){ …

int pi=3.14;area= cal_area(r);cir = 2*pi*r;

}

Extract function from code

void main(){

double r,area;printf(“input radius>”);scanf(“%lf”,&r);area= 3.14*r*r;printf(“area is %lf”,area);

}

r is an inputType of r is double

Need to output as double

Link to Example of Function

B4:

B4:

Global Variables

Local Variables

B4:

Solution:

a. local variable; A variable defined inside a function

b. general variable A variable as input to the function

c. prototype variable A function prototype is a function header without implementation.

d. global variable A variable defined outside a function

B7:

B7:

What is the output?#include <stdio.h>int x,y,z;void junk(int x,int y,int z){

x++; y++; z++;printf(“%10d%10d%10d\n”,x,y,z);

}main(){

x=1; y=2; z=3;printf(“%10d%10d%10d\n”,x,y,z);junk(x,y,z);printf(“%10d%10d%10d\n”,x,y,z);

}

What is the output?#include <stdio.h>int x,y,z;void junk(int x,int y,int z){

x++; y++; z++;printf(“%10d%10d%10d\n”,x,y,z);

}main(){

x=1; y=2; z=3;printf(“%10d%10d%10d\n”,x,y,z);junk(x,y,z);printf(“%10d%10d%10d\n”,x,y,z);

}

1 2 32 3 41 2 3

What is the output?#include <stdio.h>int x,y,z;void junk(int *x,int *y,int *z){

(*x)++; (*y)++; (*z)++;printf(“%10d%10d%10d\n”,*x,*y,*z);

}main(){

x=1; y=2; z=3;printf(“%10d%10d%10d\n”,x,y,z);junk(&x,&y,&z);printf(“%10d%10d%10d\n”,x,y,z);

}

What is the output?#include <stdio.h>int x,y,z;void junk(int *x,int *y,int *z){

(*x)++; (*y)++; (*z)++;printf(“%10d%10d%10d\n”,*x,*y,*z);

}main(){

x=1; y=2; z=3;printf(“%10d%10d%10d\n”,x,y,z);junk(&x,&y,&z);printf(“%10d%10d%10d\n”,x,y,z);

}

1 2 32 3 42 3 4

Call by Reference

• We can solved problem of multiple output from function using Call by Reference

Link to Example of Call by Reference

B1:

B1:

Solution:

a. int x; creates a variable x as integer type

b. ptr x; Not found this type of declaration

c. int &x; creates a reference to an int named 'x'

d. int *x; x is a pointer to an integer

B2:

B2:

Solution:

a. *a; value of memory address of variable named 'a'

b. a; variable named 'a

c. &a; memory address of variable named 'a'

d. address(a); function named 'address'

B15:

B15:

B16:

B16:

Solution:

a. char * Example: “123”, “123abc”, ”abc”

b. float *; Example: 1.23, 1.0, 100

c. int Example: 1, 2, 3, 10, 100, 1000

d. char Example: ‘a’, ‘b’, ‘1’, ‘2’, ‘3’

B17:

B17:

Solution:

a. char *s=“Test”

b. char s[5]={'t','e','s','t',0};

c. char s[]=“Hello”

d. char s[5]=“Hello”

T E S T \0

T E S T \0

H e l l o \0

H e l l o !! @

B20:

B20:

2

p

3 4 51

a. p=a+3;

2

p

3 4 51

c. p++;

2 3 4 51d. *q=(*p);

P

q

B3:

B3:

Solution:

a. *p; value of the variable pointed to by pointer p

b. p; memory address of the variable pointed to by

pointer p

c. &p; memory address of pointer p

d. address(p); function named 'address'

B12:

B12:


Link to Solution


Result: x = 32 y = 20

StructureStructures are the C equivalent of records. A

structure type can define in 2 ways, which arestruct Foo{

int x;int array[100];

};Usage: struct Foo f; f.x = 54; f.array[3]=9;

typedef struct{

int x;int array[100];

} Foo;Usage: Foo f; f.x = 54; f.array[3]=9;

Link to Example of Structure

StructurePassing Structs as ParametersStructs are passed by value, just like primitives void func(struct Foo foo){ foo.x = 56; foo.array[3]=55; } void main(){ struct Foo f; f.x = 54; f.array[3]=9; func(f); printf("%d and %d",f.x,f.array[3]); }

StructurePassing Structs as ParametersTo have changes occur, send a pointer to the struct void func(struct Foo* foo){ (*foo).x = 56; (*foo).array[3]=55; } void main(){ struct Foo f; f.x = 54; f.array[3]=9; func(&f); printf("%d and %d",f.x,f.array[3]); }

What will be the output of the following code and why?

#include <stdio.h>typedef struct {

char *a, *b, *c;} colors;void funct(colors sample){

sample.a = “Cyan”; sample.b = “Brown”; sample.c = “Yellow”;printf(“%s %s %s\n”, sample.a,sample.b,sample.c);

}main() {

colors sample = {“red”, “green”, “blue”}; printf(“%s %s %s\n”, sample.a,sample.b,sample.c);funct(sample); printf(“%s %s %s\n”, sample.a,sample.b,sample.c);

}

What will be the output of the following code and why?




}main() {


}

red green blueCyan Brown Yellowred green blue

If we want output as followed, how to change the program?




}main() {


}

red green blueCyan Brown YellowCyan Brown Yellow



char *a, *b, *c;} colors;void funct(colors *sample){


}main() {


}




char *a, *b, *c;} colors;void funct(colors *sample){


}main() {

colors sample = {“red”, “green”, “blue”}; printf(“%s %s %s\n”, sample.a,sample.b,sample.c);funct(&sample); printf(“%s %s %s\n”, sample.a,sample.b,sample.c);

}




char *a, *b, *c;} colors;void funct(colors *sample){ (*sample).a = “Cyan”; (*sample).b = “Brown”; (*sample).c = “Yellow”; printf(“%s %s %s\n”, sample.a,sample.b,sample.c);}main() {

colors sample = {“red”, “green”, “blue”}; printf(“%s %s %s\n”, sample.a,sample.b,sample.c);funct(&sample); printf(“%s %s %s\n”, sample.a,sample.b,sample.c);

}


B27:

B8:

C++ Spring 2000 Arrays

typedef struct tnode {int a,b,c,d;} node;main() {node x; int *msg = &x;}

2

msg

3 4 5x.a x.b x.c x.d

B8:

C++ Spring 2000 Arrays

typedef struct tnode {int a,b,c,d;} node;main() {node x; int *msg = &x;}

2

msg

3 4 5x.a x.b x.c x.d

4 Bytes

Use the given part of the program to answer questions 28-30

B28:

B28:

B29:

4 Byte40 Byte118 Byte

4 Byte

B29:

4 Byte40 Byte118 Byte

4 Byte

B30:

B30:

Wrong type argument

NULL undeclared


Link to Solution


Result: 8 6 6

B19:

Solution:

a. fseek sets the file position indicator for the stream pointed to by stream.

b. fscanf read data from file into specified variable.

c. rewind Moving the file position marker back to the beginning of the file.

d. feof Check whether the end-of-file marker is reached.

B19:

Solution:

a. fseek sets the file position indicator for the stream pointed to by stream.

b. fscanf read data from file into specified variable.

c. rewind Moving the file position marker back to the beginning of the file.

d. feof Check whether the end-of-file marker is reached.

B23:

B23:

B23:

B24:

B24:

B25:

B25:

B26:

B26:

While Loop Structure while (condition) { …. // This line will process when while condition is true }

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