experimenting... "to understand pointers more"
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this is my code: #include using namespace std; char *pchar[6]; char character[6] = "hello"; int main() { strcpy(pchar, &character); return 0; } that is "all" my code, no more no less. the problem accures on line: strcpy(pchar, &character); im not very sure on what this'll do, what i am trying to do is, have pchar point to the address of character. i thought that if there was more than one character i should use strcpy(), but im not very sure. how would i accomplish this? "see any problems, or better way of doing things, just note it :)" Thanks! ~SilverShalkin
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this is my code: #include using namespace std; char *pchar[6]; char character[6] = "hello"; int main() { strcpy(pchar, &character); return 0; } that is "all" my code, no more no less. the problem accures on line: strcpy(pchar, &character); im not very sure on what this'll do, what i am trying to do is, have pchar point to the address of character. i thought that if there was more than one character i should use strcpy(), but im not very sure. how would i accomplish this? "see any problems, or better way of doing things, just note it :)" Thanks! ~SilverShalkin
Two problems: 1) pchar is an array of char*, not char. So you will not be able to access the elements in the pchar array as a string. Which means that you will not be able to send that array to strcpy. What you would actually do with an array like pchar is point to 6 different strings. Then when you wanted to use one of the strings you would access the string in the array like so: strcpy(pchar[0], ...); 2) character is a char array, which is equivalent to one char*. Therefore you do not need to take the address of the character array, you could directly pass it into strcpy like this: strcpy(..., character); Good Luck
Build a man a fire, and he will be warm for a day
Light a man on fire, and he will be warm for the rest of his life! -
Two problems: 1) pchar is an array of char*, not char. So you will not be able to access the elements in the pchar array as a string. Which means that you will not be able to send that array to strcpy. What you would actually do with an array like pchar is point to 6 different strings. Then when you wanted to use one of the strings you would access the string in the array like so: strcpy(pchar[0], ...); 2) character is a char array, which is equivalent to one char*. Therefore you do not need to take the address of the character array, you could directly pass it into strcpy like this: strcpy(..., character); Good Luck
Build a man a fire, and he will be warm for a day
Light a man on fire, and he will be warm for the rest of his life!but i want to point to character, if i dake the address (&) off, it wouldnt be point to character would it? so i would have to declare my pchar like char *pchar[6]; pchar[0] = &character[0]; pchar[1] = &character[1]; pchar[2] = &character[2]; pchar[3] = &character[3]; pchar[4] = &character[4]; pchar[5] = &character[5]; ? i hope thats wrong... thats allot to type for somthing so small :( Thank! ~SilverShalkin :rose:
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this is my code: #include using namespace std; char *pchar[6]; char character[6] = "hello"; int main() { strcpy(pchar, &character); return 0; } that is "all" my code, no more no less. the problem accures on line: strcpy(pchar, &character); im not very sure on what this'll do, what i am trying to do is, have pchar point to the address of character. i thought that if there was more than one character i should use strcpy(), but im not very sure. how would i accomplish this? "see any problems, or better way of doing things, just note it :)" Thanks! ~SilverShalkin
Do this instead :- char *pchar; char abc[6]; strcpy(abc,"hello"); pchar = abc; Nish
Regards, Nish Native CPian. Born and brought up on CP. With the CP blood in him.
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but i want to point to character, if i dake the address (&) off, it wouldnt be point to character would it? so i would have to declare my pchar like char *pchar[6]; pchar[0] = &character[0]; pchar[1] = &character[1]; pchar[2] = &character[2]; pchar[3] = &character[3]; pchar[4] = &character[4]; pchar[5] = &character[5]; ? i hope thats wrong... thats allot to type for somthing so small :( Thank! ~SilverShalkin :rose:
Lets assume for a minute that you did do that. You could then do this: printf("%s\n",pchar[0]); printf("%s\n",pchar[1]); printf("%s\n",pchar[2]); printf("%s\n",pchar[3]); printf("%s\n",pchar[4]); printf("%s\n",pchar[5]); and the results would be: hello ello llo lo o <> in summary an array of characters IS a pointer to a sequence of characters. so if we were to do this instead char characters[6] = "hello"; char* pchar = characters; char* pchars = &characters[0]; then examine the values of all three variables in the debugger. they would all be identical. Hope this helps in your struggle with pointers. Roger.
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Do this instead :- char *pchar; char abc[6]; strcpy(abc,"hello"); pchar = abc; Nish
Regards, Nish Native CPian. Born and brought up on CP. With the CP blood in him.
would that work? - ill insert some of the code replies and try to understand this more... thanks guys! :) ~SilverShalkin :rose:
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would that work? - ill insert some of the code replies and try to understand this more... thanks guys! :) ~SilverShalkin :rose:
SilverShalkin wrote: would that work? Of course, it will work :-) Nish
Regards, Nish Native CPian. Born and brought up on CP. With the CP blood in him.
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this is my code: #include using namespace std; char *pchar[6]; char character[6] = "hello"; int main() { strcpy(pchar, &character); return 0; } that is "all" my code, no more no less. the problem accures on line: strcpy(pchar, &character); im not very sure on what this'll do, what i am trying to do is, have pchar point to the address of character. i thought that if there was more than one character i should use strcpy(), but im not very sure. how would i accomplish this? "see any problems, or better way of doing things, just note it :)" Thanks! ~SilverShalkin
WARNING: This is a long post. I think your real problem here is a misunderstanding of how C style pointers work and how to use them. Let me start by saying that a pointer is an intrinisic data type just like float, int, char, etc. The pointer data type (on Win32 systems) is a 32 bit value. This means that when you write a statement like
void\* pSomeValue;you are telling the compiler to allocate 32 bits of memory for your program. It also identifies that you intend to populate that variable with a memory address. In the case of a void type pointer you are saying that the memory pointed to by the variable can be of any type you want. (In other words, no type affinity.) C and C++ allows pointers to have type affinity also. This means that instead of being a void pointer, you can define a pointer like
char\* pszSomeValue;OR
long\* plSomeValue;OR
CSomeClassName\* pSomeValue;ETC. When you define a pointer like this you are telling the compiler that you want it to allocate 32 bits of storabe for a variable that will eventually point to a specific data type (char, long, CSomeClassName). The next subject of confusion (I think) is when you are using C style arrays of a particular data type. In your example, you used char* pchar[6] and char character[6] = "hello". The char* pchar[6] is instructing the compiler to allocate 6 memory locations each having a type affinity for a pointer to a char. This means that the compiler will allocate six 32 bit memory locations whose values it expects to be pointers to a char (char*). The char character[6] is instructing the compiler to allocate 6 memory locations each having a type affinity for a char (NOT a pointer to a char). This means that the compiler will allocate six 8 bit memory locations (that is the size of a char data type) whose value it expects to be a char. One of the nuiansces of C style arrays is that when you allocate a C style array the compiler guarantees that the memory it allocates will be continuous. This means that when you do char *pchar[6];, the compiler is going to allocate 32*6 bits in a row. When you do char character[6], the compiler is going to allocate 8*6 bits in a row. OK. That was a long winded explanation of pointers and arrays. With that understanding, lets look at the line of code where the compiler is giving you a problem.
strcpy(pchar, &character);The strcpy function is defined as taking 2 paramaters, a char* an
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WARNING: This is a long post. I think your real problem here is a misunderstanding of how C style pointers work and how to use them. Let me start by saying that a pointer is an intrinisic data type just like float, int, char, etc. The pointer data type (on Win32 systems) is a 32 bit value. This means that when you write a statement like
void\* pSomeValue;you are telling the compiler to allocate 32 bits of memory for your program. It also identifies that you intend to populate that variable with a memory address. In the case of a void type pointer you are saying that the memory pointed to by the variable can be of any type you want. (In other words, no type affinity.) C and C++ allows pointers to have type affinity also. This means that instead of being a void pointer, you can define a pointer like
char\* pszSomeValue;OR
long\* plSomeValue;OR
CSomeClassName\* pSomeValue;ETC. When you define a pointer like this you are telling the compiler that you want it to allocate 32 bits of storabe for a variable that will eventually point to a specific data type (char, long, CSomeClassName). The next subject of confusion (I think) is when you are using C style arrays of a particular data type. In your example, you used char* pchar[6] and char character[6] = "hello". The char* pchar[6] is instructing the compiler to allocate 6 memory locations each having a type affinity for a pointer to a char. This means that the compiler will allocate six 32 bit memory locations whose values it expects to be pointers to a char (char*). The char character[6] is instructing the compiler to allocate 6 memory locations each having a type affinity for a char (NOT a pointer to a char). This means that the compiler will allocate six 8 bit memory locations (that is the size of a char data type) whose value it expects to be a char. One of the nuiansces of C style arrays is that when you allocate a C style array the compiler guarantees that the memory it allocates will be continuous. This means that when you do char *pchar[6];, the compiler is going to allocate 32*6 bits in a row. When you do char character[6], the compiler is going to allocate 8*6 bits in a row. OK. That was a long winded explanation of pointers and arrays. With that understanding, lets look at the line of code where the compiler is giving you a problem.
strcpy(pchar, &character);The strcpy function is defined as taking 2 paramaters, a char* an
it helps allot, thank you for spending the amount of time that it took to type that long thing :) I know how pointers and stuff work, its just, there are a couple of things that are in the gray, as in (I dont know if it will do the exact thing i want it) i guess i should use strcpy() for a pointer, "it doesnt make sense to do so :)" i think i should print this out... :) Thanks for your time! ~SilverShalkin :rose: