String

Chapter - 7
• Introduction
• Representation of string
• Operation on string

• String is a primitive data structure.
• It is represented as sequence of
characters.
•“hello” = ‘h’, ’e’, ’l’, ’l’, ’o’

Define :-
Char name[20]; // 1D array
Here name is array of characters and
treated as string.
Size of name is 20

Char name[20]=“SDS”;
Char name[20]={‘S’, ’D’, ’S’,’0’};
‘S’ ‘D’ ‘S’ ‘0’ blank blank

Char name[ ]=“Diploma”;
Size will be allocated 8 (string
length+1) as ‘0’ will be counted.

Char name[ ][ ]={“mahesh”,
“suresh”, “mukesh”, “ramesh”};
‘m’ ‘a’ ‘h’ ‘e’ ‘s’ ‘h’ ‘0’
‘s’ ‘u’ ‘r’ ‘e’ s’ ‘h’ ‘0’
‘m’ ‘u’ ‘k’ ‘e’ s’ ‘h’ ‘0’
‘r’ ‘a’ ‘m’ ‘e’ s’ ‘h’ ‘0’

Char name[20]=“SDS”;
‘S’ ‘D’ ‘S’ ‘0’ blank blank
1. Initialize len =0;
2. Perform following steps until len reaches to ‘0’
i) otherwise increment value of len by1.
3. return len.
len

int strlen(char str[])
{
int len;
for(len=0;str[len] !=‘0’;len++);
return len;
}

int strlen(char str[]);
void main()
{ int length;
char name[20];
clrscr();
gets(name);
length=strlen(name);
printf("length is : %d",length);
getch();
}
int strlen(char str[])
{
int len;
for(len=0;str[len] !='0';len++);
return len;
}

Char name[20]=“SDSRK”;
‘S’ ‘D’ ‘S’ ‘R’ ‘K’ ‘0’ blank
i j

‘S’ ‘D’ ‘S’ ‘R’ ‘K’ ‘0’ blank
i j
s
k

‘K’ ‘D’ ‘S’ ‘r’ ‘S’ ‘0’ blank
i j
D R

‘K’ ‘R’ ‘S’ ‘D’ ‘S’ ‘0’ blank
ij
1. Find length of string
2. Initialize i=0 and j=length-1.
3. Perform following steps till i less than length/2
i) exchange the element at ith and jth position.
ii) increment i by 1
iii) decrement j by 1

void reverse(char str[])
{
int i,j,len;
char temp;
for(len=0;str[len]!='0';len++);
for(i=0,j=len-1;i<len/2;i++,j--)
{
temp=str[i];
str[i]=str[j];
str[j]=temp;
}
return;
}

‘S’ ‘D’ ‘S’ ‘0’ blank blank blank
‘R’ ‘K’ ‘U’ ‘0’ blank blank
i
j
R

‘S’ ‘D’ ‘S’ ‘R’ blank blank blank
i
j
K

‘S’ ‘D’ ‘S’ ‘R’ ‘K’ blank blank
i
j
U

‘S’ ‘D’ ‘S’ ‘R’ ‘K’ ‘U’ blank
i
j
‘0’

‘S’ ‘D’ ‘S’ ‘R’ ‘K’ ‘U’ ‘0’ blank
i
j

1. Find length of string1.
3. Initialize i=length of string1 and j to 0.
4. Perform following steps till j not reaches to
length of string2
i) copy jth element of string2 in ith element of
string1.
ii) increment i and j by 1.
5. Store ‘0’ to ith element of string1.

void concate(char str1[],char str2[])
{
int i,j,len1,len2;
for(len1=0;str1[len1]!='0';len1++);
for(i=len1,j=0;j<len2;j++,i++)
{
str1[i]=str2[j];
}
str1[i]='0';
}

‘s’ ‘d’ ‘s’ ‘0’ blank blank blank
i
s
Convert t to
uppercase

‘s’ ‘d’ ‘s’ ‘0’ blank blank blank
i
S Convert t to
uppercase
s

‘S’ ‘d’ ‘s’ ‘0’ blank blank blank
i
D Convert t to
uppercase
d

‘S’ ‘D’ ‘s’ ‘0’ blank blank blank
i
S Convert t to
uppercase
s

‘S’ ‘D’ ‘S’ ‘0’ blank blank blank
i

1. Find length of string.
2. Initialize i = 0.
3. Perform following steps till i is not equal to length of string.
a) check whether ith element of string is lowercase or not.
i) if it is lowercase then subtract 32 from ASCII value.
b) increment i by 1.

void convert (char str[])
{
int i,len;
for(i=0; i< len; i++)
{
if(str[i]>=‘a’ && str[i]<=‘z’)
str[i]= str[i]-32;
}
}

Char name1[ ]=“SDSRK”,name2[ ] = “SDSKK”;
‘S’ ‘D’ ‘S’ ‘R’ ‘K’ ‘0’ blank
i
‘S’ ‘D’ ‘S’ ‘K’ ‘K’ ‘0’ blank
mismatch

1. Initialize i=0.
2. Perform following steps till any of the strings
get ‘0’
i) check ith element of both the strings.
ii) if they are not same then return the difference
of ASCII value
iii) else increase the value of i by 1.
3. If both strings are same, return 0.

int compare(char str1[],char str2[])
{
int i;
for(i=0;str1[i]!='0'||str2[i]!='0';i++)
{
if (str1[i]!=str2[i])
return (str1[i]-str2[i]);
}
return 0;
}

‘s’ ‘d’ ‘s’ ‘r’ ‘k’ ‘0’ blank
i
‘r’ ‘k’ ‘0’ blank blank blank
j

3. Initialize i = 0, j=0.
4. Perform following steps till i is less or equal to difference of
strings.
a) check whether ith element of string1 is equal to jth element of
string2.
b) if they are same then check other elements of string2 with
string1 untill length of string2
c) increment i by 1.
5. If substring is not found then return 0.

int substring(char str1[],char str2[])
{
int i,j,k,len1,len2;
for(i=0,j=0;i<=len1-len2;i++)
{ if (str1[i]==str2[j])
{
for(j=0,k=i;j<len2;k++,j++)
{ if(str1[k]!=str2[j])
return 0;
}
return i;
}
}
return 0;
}

‘m’ ‘i’ ‘n’ ‘i’ ‘m’ ‘0’

1. Find length of string.
2. Initialize i to 0 and j to length-1.
3. Perform following steps till i is less than half of string.
a) check whether ith element and jth element are same or not.
b) if not than return ASCII difference
c) increment i by 1 and decrement j by 1.
5. return 0.

int palindrome(char str[])
{
int i,j,len;
i=0;j=len;
while(i<len/2)
{
if(str[i]!=str[j])
return str[i]-str[j];
i++; j--;
}
return 0;
}

int palindrome(char str[]) {…………..}
void main()
{
char str[20];
int s;
clrscr();
gets(str);
s=palindrome(str);
printf("nString is %s palindrome.",(s==0)?("a"):(s));
getch();
}

HW
Write an algorithm and function that gives ascii difference of 2 characters.
Ie. String is ‘abcd’
Output:
a-b =ASCII (1)
b-c=ASCII (1)
c-d=ASCII (1)

String

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