I am only just learning pointers and finding them quite difficult. When I try to compile I get pointers.c:18: error: expected ‘;’, ‘,’ or ‘)’ before ‘&’ token. I am not to sure what this means.
Is it ok for me to do this?
int alpha;
getAlpha(*alpha);
void getAlpha(int &alpha)
.... rest of code.
Sorry if this doesn't make sense
No, you have your * and your & mixed up.
Please use code tags next time.
Code:void getAlpha(int *alpha) { *alpha = 5; } void foo(void) { int alpha; getAlpha(&alpha); }
If you understand what you're doing, you're not learning anything.
(use code tags)
I dont understand exactly what you are trying to do...
But if you want to pass pointer as argument to your function, then let me explain:
Sorry for being pedanticCode:int getAlpha(int *alpha) { *alpha = 1; return 0; } int main( void ) { int alpha = 0; if( !getAlpha( &alpha ) ) { printf( "Got alpha: %d\n", alpha ); } else { printf( "Error getting alpha!\n" ); } return 0; }
You initialize variable alpha which is just an int variable. Then with &alpha you get it's pointer (reference) which you can pass to getAlpha() function as argument since it requires pointer *.
In function we assign pointer * alpha int value 1, so we need to dereference the pointer: *alpha = 1;
And about me being pedantic, every function should return value indicating it's success and caller function should handle that return value, in one way or another. Just couldnt write simple example without doing it this way.
So, int he basics; &var is the reference to var. *var is dereferenced var.
That's why if you have int and you want to make pointer pointing to it, you type:
So, pointer pvar points to variable var. Reference & is almost the same thing as pointer to it's variable. So, it's possible to use variable var = 1 or *&var = 1 since * dereferences the & reference.Code:int var = 0; int *pvar = &var; // OR int *pvar; pvar = &var; // then use pointer to change var's value from 0 to 1: *pvar = 1;
Spooky aint it?
Last edited by Rennor; 08-24-2006 at 11:35 PM.
Sorry I forgot to put the tags on!
Thanks, I will have another try now and see how it goes.
No, in your getAlpha() you are trying to de-reference a pointer but you did not decalre alpha as a pointer.
A simple example may help -
<edit> wow, you guys are fast, nearly half a dozen posts while I was writing mine!Code:#include <stdio.h> /* declare your function*/ void getAlpha(int &alpha){ printf("Alpha's value = %d \n",alpha); } int main(int argc, char* argv[]) { /* decalre and initialise a normal int var.*/ int alphavalue = 6; /* declare a pointer that points to an int (holds its address) but currently holds nothing and will crash your app if you don't point it to something before using it!*/ int * alpha = NULL; /*pass the address of the normal int, now the pointer actually points to something, very important!*/ alpha = &alphavalue; /* use your function, now you can dreference your pointer */ getAlpha(*alpha); return 0; }
I'm still playing with it but I just can't work it out sorry.
This all the code. Ignore the logic errors I'm still playing with those.
Thanks guys.Code:#include <stdio.h> int main() { int alpha; int *alpha = NULL; alpha = &alpha getalpha(*alpha); if (ispointers(alpha)) printf("\n%d COMMENT HERE\n", alpha); else printf("\n%d OPP COMMENT HERE\n", alpha); return 0; } void getalpha(int &alpha) { printf("INSTRUCTION "); if (scanf("%d", alpha) != 2 || *alpha < 0) { printf("Invalid number entered\n"); exit(1); } } int ispointers(int alpha) { int pal, temp = alpha; while (temp > 0) { pal = pal / 10 + temp % 10; temp /= 10; } return (pal = alpha); }
int alpha and int* alpha are the same (or close enough to confuse the compiler when you use alpha somewhere else) so you can as a identifier use a p in from of your pointer def's like so
int* pAlpha;
How many alphas are you trying to declare exactly? variables in the same scope have to have unique names. Maybe call the pointer int *palpha or something.Code:int alpha; int *alpha = NULL; alpha = &alpha
If you understand what you're doing, you're not learning anything.
Ahhh ok I've fixed it up. I think I understand it now.
Thanks alot for you're help. (",)
