c pointer with string

Trying to learn pointers be开发者_开发问答tter I wrote this code. The intention was to print one character of a string at a time in a for loop. Again trying to learn how to increment a pointer and then dereference the pointer.

char *string = "abcdef";
char *pointer = &string; 

for (int i =0; i < 4; i++)

{   
    printf("the next charater is %c\n", *pointer);
    pointer = (char *)pointer + sizeof(char);
}

want it to print:

the next charater is a

the next charater is b

the next charater is c

the next charater is d

char *pointer = &string; 

should be

char *pointer = string; 

string is a pointer variable that contains the address of your string literal. You want the address of the string literal, so you should simply copy the value in string - not take the address of the local pointer variable - your current code gives you a pointer to a pointer to a string.

Additionally, pointer = (char *)pointer + sizeof(char); doesn't require a cast, and it should not use sizeof(char). When incrementing a variable of type pointer to X, incrementing it by one will increment it by sizeof(X) bytes - increment by one to point at the next X. Use pointer += 1; or ++pointer instead.

If you want to print (or otherwise process) the whole string one char at a time, you may use the following idiom:

char *p = /* ... */;

while (p && *p) {
    printf("next char: %c\n", *p++);
}

The condition first tests whether p is NULL, i.e. if it is wise to dereference p at all. If p is not NULL, *p tests whether you already reached the end of the string, denoted by the '\0' character, which happens to be 0.

If you want to use pointer to pointer to string then you can use like this:

char *string = "abcdef";
char **pointer = &string;
int i;
for (i =0; i < 4; i++)

{
    printf("the next charater is %c\n", **pointer);
    *pointer = *pointer + sizeof(char);
}