A string literal is a sequence of characters terminated by null byte ‘0’. For example,
/* strlit_op.c -- program shows operations on string literals */ #include <stdio.h> #define NAME "What is your name?" int main(void) { char name[25] = "xyz"; char * const p2name = "abc"; "abc" += 1; "abc" + 1; "abc"[0]; "abc"[1]; "abc"[10]; *("abc" + 2); *"abc"; return 0; }
If you compile the above program, compiler saves copy of #define sting some where in the memory and set a pointer to it. Compiler also terminates the string with null terminator. What about name character array? It’s a pointer constant. And p2name? It’s also a pointer constant. Let’s try to figure out how expressions in above program evaluate to,
"abc" += 1;
It seems as if we add 1 to string literal “abc” and assigns it the result. What do you think, if you can figure it out. This statement is an error. Because value of string literal is a pointer constant. Like, for ex., exp.,
5 += 5; /* 5 is constant */
can’t be evaluated. Therefore, statement
"abc" += 1;
is invalid. Consider the following statement,
"abc" + 1;
Again we are trying to add 1 to the string literal “abc”. Sure, you can, this time, easily unravel this. Since, string literal “abc” is a pointer constant pointing to first character ‘a’. Type of “abc” is a pointer to char. Therefore, when you add 1 to “abc”, 1 is adjusted to type char * and added to pointer constant. This moves pointer to character ‘b’.
Take another exp.,
"abc"[0];
As we know that array name is a pointer constant pointing to first element of the array. Therefore, exp. “abc”[0] can be rewritten as
*("abc" + 0);
or even as
*"abc";
which results first character ‘a’ in the string “abc”. Let’s explore exp.
"abc"[10];
Let’s take an analogy to evaluate the above exp.,
int white[5]; /* white is an array of 5 integers */
What will happen if you attempt to access
white[10];
Surely, subscript value 10 takes you off the end of the array white. Hence, an undefined result. Similarly, in exp.,
"abc"[10];
you go out of bounds and therefore undefined result.
/* strlit_op.c -- program shows operations on string literals */ #include <stdio.h> #define NAME "What is your name?" int main(void) { char name[25] = "xyz"; char * const p2name = "abc"; //"abc" += 1; printf("Result of \"abc\" + 1 is %c\n", "abc" + 1); printf("Result of *(\"abc\" + 1) is %c\n", *("abc" + 1)); printf("Result of \"abc\"[0] is %c\n", "abc"[0]); printf("Result of \"abc\"[1] is %c\n", "abc"[1]); printf("Result of \"abc\"[10] is %c\n", "abc"[10]); printf("Result of *(\"abc\" + 2) is %c\n", *("abc" + 2)); printf("Result of *\"abc\" is %c\n", *"abc"); return 0; } Output of the above program is as follows, <pre lang="C" cssfile="hk1_style"> Result of "abc" + 1 is Result of *("abc" + 1) is b Result of "abc"[0] is a Result of "abc"[1] is b Result of "abc"[10] is Result of *("abc" + 2) is c Result of *"abc" is a
Let’s see an application of string literal in a C program that converts an unsigned integer into equivalent hexadecimal representation,
/* * strlit_app.c -- program converts unsigned values into hexadecimal * number */ #include <stdio.h> #define MINUS1 4294967295 /* unsigned representation of -1 */ /* uint_to_hex() is a recursive function */ void uint_to_hex(unsigned); int main(void) { unsigned value; puts("\n**Program converts Unsigned int to Hexadecimal No.**\n"); printf("User, enter an unsigned integer value or -1 to quit: "); while (scanf("%u", &value) == 1 && value != MINUS1) { /* call to uint_to_hex() */ uint_to_hex(value); puts(""); printf("more conversions, enter unsigned integer or -1 to quit:"); } puts("Thank You!"); return 0; } void uint_to_hex(unsigned value) { unsigned quotient; quotient = value / 16; if (quotient != 0) uint_to_hex(quotient); putchar("0123456789ABCDEF"[value % 16]); }
Output of the program as follows,
**Program converts Unsigned int to Hexadecimal No.** User, enter an unsigned integer value or -1 to quit: 54 36 more conversions, enter unsigned integer or -1 to quit: 12 C more conversions, enter unsigned integer or -1 to quit: 8667 21DB more conversions, enter unsigned integer or -1 to quit: 12 C more conversions, enter unsigned integer or -1 to quit: 3232 CA0 more conversions, enter unsigned integer or -1 to quit: -1 Thank You!
Notice how easy it has become to convert unsigned values to hex exploiting string literal. Exp.
[value % 16]
in square parenthesis evaluates subscript in the exp.
"0123456789ABCDEF"[value % 16]
and then selected the corresponding character from the string before printed that.
Sanfoundry Global Education & Learning Series – 1000 C Tutorials.