hi,
i have this code:
Code:#include <stdio.h> #include <string.h> #include <stdlib.h> int main(int argc, char **argv) { int *p; char *ddr = "abc"; *p=ddr; }
what i want is bassicaly to assign to the *p the pointer of the string so that
i could do the following printf(" print string %s\n",*p);
so i dont know how to do that.
i thank you in advance for the help
I can tell you right now that's impossible. Your compiler will give you an warning saying that the %s format expects a 'char *', but you gave it a value of type int. Then if you run your program, heaven forbid, your program will crash.
If you really want to see how to convert a character pointer to an integer pointer, here is a working example:
Code:#include <stdio.h> #include <stdlib.h> int main( ) { int * i = NULL; char * secret = "This is a string literal!"; i = ( int * )secret; printf( "%d\n", *i ); return 0; }
Last edited by MilleniumFalcon; 03-22-2014 at 04:22 PM.
i am compiling with the following options: gcc -g -Wall -ansi -pedantic testes.c -o tt
It doesn't matter, it will still crash because you tried to print an integer like a string, which just doesn't work. If you're wanting to do something different, ask.Originally Posted by tindala
Why are you trying to store a char * in an int, anyway?
I mean, it's not impossible to cast an address (which is what a pointer is, anyway) into an integer, but in general you have no guarantee that sizeof(int) is greater or equal to sizeof(char *), so you might overflow.
You ever try a pink golf ball, Wally? Why, the wind shear on a pink ball alone can take the head clean off a 90 pound midget at 300 yards.
If you want to see what is in the pointer, usewhat i want is bassicaly to assign to the *p the pointer of the string so that
i could do the following printf(" print string %s\n",*p);
Code:char *ddr = "abc"; printf("Address is: %p", add);
Fact - Beethoven wrote his first symphony in C
What output do you expect to get when you do that?
Note: as others have said, what you've described is a REALLY bad idea.
