How are character pointers handled in C?

Answer 1

In C, pointers are addresses to specific locations in memory. They contain no additional information. It's up to the compiler, memory management system or programmer to keep track of how much space is being used by the program that allocates the memory.

C allows programmers to assign a type to the variable containing the pointer address. In the case of this question, the pointer variable references memory in one-byte chunks, each byte pertaining to a 8-bit signed integer called a "character", or char.

Also keep in mind that pointers are nothing more than glorified integers. In modern C Programming, these integers are 32-bit values. Since integers can be added to and subtracted from, pointers can as well, and the following code is perfectly legitimate:

#include

#include

void main()

{

char *mystr="Hello there!";

printf("ccc%c",

*mystr, *(mystr+4), *(mystr+3), *(mystr+2), *(mystr+8), *(mystr+9), *(mystr+strlen(mystr)-1));

}

By prefixing the pointer variable with an asterisk (*), the code is telling the compiler to access the value at the memory location specified by a variable by the type of the variable. *mystrrefers to the character located at the address of the literal string "Hello there!", which is an 'H'.

The above code is also referencing memory location offsets from the address in mystr. mystr+4 refers to the memory location 4 bytes past mystr's address, which is the 'o' in "Hello".

The last wonderfully messy parameter adds the length of the string data at mystr to the address of mystr and subtracts 1. Since strings in C are NUL-terminated (zero-terminated), adding the length of the string to the memory location of the string refers to the value 0 terminating the string. Subtracting 1 results in the last character of the string, which in this case is the exclamation mark.

All pointer variables can also point to memory addresses returned by the malloc() function. For instance, here's a simple string duplication function like strdup():

#include

char *stringdup(char *str)

{

char *returnstr;

returnstr=malloc(strlen(str)+1);

memcpy(returnstr, str, strlen(str)+1);

return returnstr;

}

Once the pointer returned from stringdup() is no longer needed, simply use free() to release the memory. This helps to avoid memory leaks in your program.

Note that, when or if you decide to begin exploring C++, you'll need to handle typecasting more intensively. C++ distinguishes between const char* and char*, for instance, and char* and void*. (malloc() returns a void*, so the stringdup() function above would produce a warning without typecasting the malloc() return value to char* when assigning it to returnstr.)

That's really all there is to know about character pointers in C.