pointer printing
I would like to print the third element of the array with pointers but I am unable to do it... even with simple address operatorCode:#include <stdio.h> main(){ int a[20]; int i; for (i=0; i<20; i=i+2) { a[i]=i; } for (i=0; i<20; i=i+2) { printf("%d", a[i]); } int *p; p=a; int dd; dd=*(p+3); int gg; gg=a[3]; printf("\n %d", dd); printf("\n %d", gg); }why?
Try printf with %p instead of %d.
Edit: Just read again what you were ACTUALLY trying to do. Sorry.
So, if you want to print the address of the third element use what I indicated above.
If you want to print the value contained at that address use %d.
NOTE, that your implementation leaves all uneven indexes of the array uninitialized. Only a[0],a[2],a[4]... will have correct values, so a[3] will probably contain garbage.
Last edited by claudiu; 06-10-2010 at 02:50 AM.
1. Get rid of gets(). Never ever ever use it again. Replace it with fgets() and use that instead.
2. Get rid of void main and replace it with int main(void) and return 0 at the end of the function.
3. Get rid of conio.h and other antiquated DOS crap headers.
4. Don't cast the return value of malloc, even if you always always always make sure that stdlib.h is included.
...What?Originally Posted by claudiu
Define "not working". Also note that the 3rd element is index 2.
And make your main function return int.
Just edited my post, I thought she was trying to print the address of the third element.
1. Get rid of gets(). Never ever ever use it again. Replace it with fgets() and use that instead.
2. Get rid of void main and replace it with int main(void) and return 0 at the end of the function.
3. Get rid of conio.h and other antiquated DOS crap headers.
4. Don't cast the return value of malloc, even if you always always always make sure that stdlib.h is included.
Why I have bad access here? (EXC_BAD_ACCESS) --> printf("\n %d", *p);Code:#include <stdio.h> main(){ int a[5]={1,4,7,10,13}; int i; printf("\n %d \n", a[10]); for (i=0; i<5; i++) { printf("%d", a[i]); } int *p; int *q; p=a; p=*(a+2); printf("\n %d", *p); q=*(a+3); if (p<q) { printf("\n TRUE"); } }
If I write like this no problem printf("\n %p", p);
please let me know
Well, clearly, a and *(a+2) are two different types no? So you are assigning to type int* two different types. One of them is wrong, I'll leave it up to you to figure out which one.
Also, p<q compares two addresses, the one pointed to by p and the one pointed to by q. Not exactly what you want is it?
Remember:
A POINTER IS AN ADDRESS. The dereferencing operation on a pointer type (i.e. *ptr) is the VALUE contained AT THE ADDRESS the pointer POINTS TO.
Last edited by claudiu; 06-10-2010 at 03:40 AM.
1. Get rid of gets(). Never ever ever use it again. Replace it with fgets() and use that instead.
2. Get rid of void main and replace it with int main(void) and return 0 at the end of the function.
3. Get rid of conio.h and other antiquated DOS crap headers.
4. Don't cast the return value of malloc, even if you always always always make sure that stdlib.h is included.
Main must return int.
Thank you.
Elysia --> what do you mean with "Main must return int"? return 0 ... or?
main doesn't have a return type in your code. It must return int.
I think main() is allowed in C89, however you are right in that everyone should adhere to the new standard.
1. Get rid of gets(). Never ever ever use it again. Replace it with fgets() and use that instead.
2. Get rid of void main and replace it with int main(void) and return 0 at the end of the function.
3. Get rid of conio.h and other antiquated DOS crap headers.
4. Don't cast the return value of malloc, even if you always always always make sure that stdlib.h is included.
It's considered bad practice to use implicit return types. C99 goes as far as banning them altogether.
Therefore, don't use them. That's why I tend to tell newbies that they must make main return int.
