Strings
In C programming, a string is a sequence of characters terminated with a
null character \0 . For example:
char c[] = "c string";
When the compiler encounters a sequence of characters enclosed in the
double quotation marks, it appends a null character \0 at the end by
default.
Memory Diagram
How to declare a string?
Here's how you can declare strings:
char s[5];
String Declaration in C
Here, we have declared a string of 5 characters.
,How to initialize strings?
You can initialize strings in a number of ways.
char c[] = "abcd";
char c[50] = "abcd";
char c[] = {'a', 'b', 'c', 'd', '\0'};
char c[5] = {'a', 'b', 'c', 'd', '\0'};
String Initialization in C
Let's take another example:
char c[5] = "abcde";
Here, we are trying to assign 6 characters (the last character is '\0' ) to
a char array having 5 characters. This is bad and you should never do this.
Assigning Values to Strings
Arrays and strings do not support the assignment operator once it is
declared. For example,
char c[100];
c = "C programming"; // Error! array type is not assignable.
Note: Use the strcpy() function to copy the string instead.
, Read String from the user
You can use the scanf() function to read a string.
The scanf() function reads the sequence of characters until it
encounters whitespace (space, newline, tab, etc.).
Example 1: scanf() to read a string
#include <stdio.h>
int main()
{
char name[20];
printf("Enter name: ");
scanf("%s", name);
printf("Your name is %s.", name);
return 0;
}
Output
Enter name: Dennis Ritchie
Your name is Dennis.
Even though Dennis Ritchie was entered in the above program,
only "Dennis" was stored in the name string. It's because there was a space
after Dennis .
Also notice that we have used the code name instead of &name with scanf() .
scanf("%s", name);
This is because name is a char array, and we know that array names decay
to pointers in C.
Thus, the name in scanf() already points to the address of the first element in
the string, which is why we don't need to use & .
In C programming, a string is a sequence of characters terminated with a
null character \0 . For example:
char c[] = "c string";
When the compiler encounters a sequence of characters enclosed in the
double quotation marks, it appends a null character \0 at the end by
default.
Memory Diagram
How to declare a string?
Here's how you can declare strings:
char s[5];
String Declaration in C
Here, we have declared a string of 5 characters.
,How to initialize strings?
You can initialize strings in a number of ways.
char c[] = "abcd";
char c[50] = "abcd";
char c[] = {'a', 'b', 'c', 'd', '\0'};
char c[5] = {'a', 'b', 'c', 'd', '\0'};
String Initialization in C
Let's take another example:
char c[5] = "abcde";
Here, we are trying to assign 6 characters (the last character is '\0' ) to
a char array having 5 characters. This is bad and you should never do this.
Assigning Values to Strings
Arrays and strings do not support the assignment operator once it is
declared. For example,
char c[100];
c = "C programming"; // Error! array type is not assignable.
Note: Use the strcpy() function to copy the string instead.
, Read String from the user
You can use the scanf() function to read a string.
The scanf() function reads the sequence of characters until it
encounters whitespace (space, newline, tab, etc.).
Example 1: scanf() to read a string
#include <stdio.h>
int main()
{
char name[20];
printf("Enter name: ");
scanf("%s", name);
printf("Your name is %s.", name);
return 0;
}
Output
Enter name: Dennis Ritchie
Your name is Dennis.
Even though Dennis Ritchie was entered in the above program,
only "Dennis" was stored in the name string. It's because there was a space
after Dennis .
Also notice that we have used the code name instead of &name with scanf() .
scanf("%s", name);
This is because name is a char array, and we know that array names decay
to pointers in C.
Thus, the name in scanf() already points to the address of the first element in
the string, which is why we don't need to use & .
