stuck at char ptr*
Author
excuse me,
why asterisk(*) sign is not required to be attached behind "s"
i read the tutorial that was quoted but i don't understand why the char ptr does not need to dereference the string
thx in advance.
quote from http://computer.howstuffworks.com/c36.htm
Fragment 1
{
char *s;
s="hello";
printf("%s\n",s);
}
"These two fragments produce the same output, but their internal behavior is quite different. In fragment 2, you cannot say s="hello";. To understand the differences, you have to understand how the string constant table works in C.
When your program is compiled, the compiler forms the object code file, which contains your machine code and a table of all the string constants declared in the program. In fragment 1, the statement s="hello"; causes s to point to the address of the string hello in the string constant table. Since this string is in the string constant table, and therefore technically a part of the executable code, you cannot modify it. You can only point to it and use it in a read-only manner.
[Edited by - mckoo on August 4, 2004 2:37:36 AM]
Quote:Original post by mckoo
excuse me,
why asterisk(*) sign is not required to be attached behind "string"
i read the tutorial that was quoted but i don't understand why the char ptr does not need to dereference the string
Because a string literal is not a pointer, you can't dereference a string literal because it's not a pointer. I think you need to read up on pointers & arrays.
A string literal is a statically allocated constant array of characters meaning it's of type
const char a[n];char a[] = { 'h', 'e', 'l', 'l', 'o', '\0' };char* ptr = a;orchar* ptr = &a[0];Except string literals in C cannot be modified otherwise it will cause undefined behaviour.
Author
sorry i made a typing mistake....
it should be why * sign cannot be added behind "s" so can be assigned with the string "hello"
it should be why * sign cannot be added behind "s" so can be assigned with the string "hello"
Quote:Original post by mckoo
sorry i made a typing mistake....
it should be why * sign cannot be added behind "s" so can be assigned with the string "hello"
Because you have already declared it as a pointer to char before, when you use the operator "*" outside a declaration of a new variable it means to dereference a pointer (to get the value it points to) so if you did that it's gonna crash because when you declared that pointer you didn't assign it to any value but thats ok because you assign it a copy of the address of first element into that string literal later on. Like i said before if you read up about pointers & arrays you'll come across a section about how the name of an array is simillar to a constant pointer to the first element in the array. Then things will start to make sense.
Author
so since the starting address is assigned to ptr....
ptr = string;
[ptr + 1] = string[1];?
----------------------------------------------
so that "ptr" could print out the content stored in the element of string?
seems making sense rite?
thx in advance~~
ptr = string;
[ptr + 1] = string[1];?
----------------------------------------------
so that "ptr" could print out the content stored in the element of string?
seems making sense rite?
thx in advance~~
Quote:Original post by mckoo
so since the starting address is assigned to ptr....
ptr = string;
[ptr + 1] = string[1];?
You mean is (ptr + 1) equivalent to string[1] yes thats correct.
Author
thx so much....
but how could i print out a ptr with string
since the address of first element of the array is only assigned to ptr
printf(" %s", ptr);
%s means that it will look up the next element of the ptr of the array?
excuse me ignorance
but how could i print out a ptr with string
since the address of first element of the array is only assigned to ptr
printf(" %s", ptr);
%s means that it will look up the next element of the ptr of the array?
excuse me ignorance
Quote:Original post by mckoo
thx so much....
but how could i print out a ptr with string
since the address of first element of the array is only assigned to ptr
printf(" %s", ptr);
%s means that it will look up the next element of the ptr of the array?
excuse me ignorance
the conversion specifier s means a null-terminated strings (C style strings) so it knows that it's either a pointer points to the first element of an array of characters, or it's an array of characters because you can use an array name just like a pointer to the first element of an array.
So inside printf you can imagine a loop through the array to print the string to the console.
This is a good example of how a pointer to an array & how the name of an array can be used like each other, take the library function strcpy (string copy) you could do it like this:
first with array subcript version
second version pointer arithmetic version
You can see how the pointer version can reduce to the common C idolism:
What is the advantage of using a pointer as parameter instead of an array? well the pointer doesn't have to point to the first element of an array, it could be in the middle.
[Edited by - snk_kid on August 4, 2004 4:17:43 AM]
first with array subcript version
void strcpy(char* source, char* dest) { int i = 0; while((source = dest) != '\0') ++i;}second version pointer arithmetic version
void strcpy(char* source, char* dest) { while((*source = *dest) != '\0') { ++source; ++dest; }}You can see how the pointer version can reduce to the common C idolism:
void strcpy(char* source, char* dest) { while((*source++ = *dest++) != '\0') ;}What is the advantage of using a pointer as parameter instead of an array? well the pointer doesn't have to point to the first element of an array, it could be in the middle.
[Edited by - snk_kid on August 4, 2004 4:17:43 AM]
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