A string in C can be addressed with a character pointer or a character array.
Strings as character arrays :
char str[4] = "LTH"; /*One extra for string terminator*/
/* OR */
char str[4] = {'L', 'T', 'H', '\0'}; /* '\0' is string terminator */When strings are declared as character arrays, they are stored like other types of arrays in C.
If str[] is an auto variable, it is put in the stack segment; if it is a global or static variable, it is saved in the data segment.
Strings using character pointers:
Using character pointer strings can be stored in two ways:
1) Read-only string in a shared segment:
When a string value is directly assigned to a pointer, in most of the compilers, it’s stored in a read-only block (generally in the data segment) that is shared among functions.
char *str = "LTH";
In the above, code “LTH” is stored in a shared-read only location but pointer str is stored in read-write memory.
You can change str to point something else but cannot change the value at present str.
So this kind of string should only be used when we don’t want to modify the string at a later stage in the program.
2) Dynamically allocated in heap segment:
Strings are stored like other dynamically allocated things in C and can be shared among functions.
char *str; int size = 4; /*one extra for '\0'*/ str = (char *)malloc(sizeof(char)*size); *(str+0) = 'L'; *(str+1) = 'T'; *(str+2) = 'H'; *(str+3) = '\0';
Let’s see some examples of a better understanding of the above ways to store string.
Example 1 (Try to modify string):
The below program may crash (gives segmentation fault error) because the line *(str+1) = ‘n’ tries to write a read-only memory.
int main()
{
char *str;
str = "LTH"; /* Stored in read only part of data segment */
*(str+1) = 'n'; /* Problem: trying to modify read only memory */
getchar();
return 0;
}The below program works perfectly fine as str[] is stored in a writable stack segment.
int main()
{
char str[] = "LTH"; /* Stored in stack segment like other auto variables */
*(str+1) = 'n'; /* No problem: String is now GnG */
getchar();
return 0;
}The below program also works perfectly fine as data at str is stored in the writable heap segment.
int main()
{
int size = 4;
/* Stored in heap segment like other dynamically allocated things */
char *str = (char *)malloc(sizeof(char)*size);
*(str+0) = 'L';
*(str+1) = 'T';
*(str+2) = 'H';
*(str+3) = '\0';
*(str+1) = 'n'; /* No problem: String is now GnG */
getchar();
return 0;
} Example 2 (Try to return string from a function):
The below program works perfectly fine as the string is stored in a shared segment and data stored remains there even after the return of getString().
char *getString()
{
char *str = "LTH"; /* Stored in read only part of shared segment */
/* No problem: remains at address str after getString() returns*/
return str;
}
int main()
{
printf("%s", getString());
getchar();
return 0;
}The below program also works perfectly fine as the string is stored in heap segment and data stored in heap segment persists even after the return of getString().
char *getString()
{
int size = 4;
char *str = (char *)malloc(sizeof(char)*size); /*Stored in heap segment*/
*(str+0) = 'L';
*(str+1) = 'T';
*(str+2) = 'H';
*(str+3) = '\0';
/* No problem: string remains at str after getString() returns */
return str;
}
int main()
{
printf("%s", getString());
getchar();
return 0;
}But, the below program may print some garbage data as the string is stored in the stack frame of function getString() and data may not be there after getString() returns.
char *getString()
{
char str[] = "LTH"; /* Stored in stack segment */
/* Problem: string may not be present after getString() returns */
/* Problem can be solved if write static before char, i.e. static char str[] = "GfG";*/
return str;
}
int main()
{
printf("%s", getString());
getchar();
return 0;
}Please write comments if you find anything incorrect in the above article, or you want to share more information about the storage of strings.
