Following is my code:
#include <stdio.h>
int main()
{
char a[10]="Hi!";
void f(char b[])
{
// 1. printf("*a is:%s\n",*a);
printf("&a is:%p\n",&a);
// 2. printf("&&a is:%p\n",&(&a));
printf("a's address is:%p\n",a);
printf("a's value is:%s\n",a);
printf("b's address is:%p\n",b);
printf("b's value is:%s\n",b);
// 3. printf("*b is:%s\n",*b);
printf("&b is:%s\n",&b);
}
f(a);
return 1;
getch();
}
Running the above code gives the output:
&a is:0028FF1C
a's address is:0028FF1C
a's value is:Hi!
b's address is:0028FF1C
b's value is:Hi!
&b is:∟ (
In the Output:
Why are there different outputs for &a and &b;
Although a and b have same reference.
Further,
I've mentioned 3 comments by their number.
If I remove slashes one by one and execute them,I get following 2 issues:
On executing comment no. 1 & 3:
"abc.exe has stopped working."
On executing comment no. 2:
abc.c: In function 'f':
abc.c:14:32: error: lvalue required as unary '&' operand
printf("&&a is:%p\n",&(&a));
^
You copy pasted the printf() and didn't change the specifier
printf("&b is:%s\n", &b)
/* ^ should be `p' */
The first comment, when executed is causing a problem because you are using the "%s" specifier and passing the first character of a which will be interpreted as an address, try this
printf("*a is: %c\n", *a);
The third comment presents the same problem.
The second comment, you need an intermidiate pointer to do that, like
char *c = &a;
char **d = &c;
printf("%p\n", (void *) d);
although it will print the same as printf("%p\n", (void *) a); because they all have the same address, the differenece is that the pointer arithmetic will work different.
Well, you shouldn't define functions inside another function - though some compilers accept that (like gcc), it isn't portable.
printf("&b is:%s\n",&b);
Here &b is the address of b, which is the address of the char array. In effect &b is the address of the parameter.
1.) and 3.) fail because you need to pass a pointer to a string. *a, is a dereferenced pointer, so, in effect, it's the first character of the string
2.) &&a is a pointer to a pointer to a pointer to a pointer (as a is already a pointer.
Note: This is the question author's own answer, see revision 3
I understood the above issues and here's the improved code:
#include<stdio.h>
#include<string.h>
int i;
char a[11]="F*** Yeahh!"; //String : char *
//'a' gets memory in "globals" area;'a' is a pointer to 'F'
void func();//Function Definition
int main()
{
func(a); //Passing reference of 'a' to func
getch();
return 1;
}
void func(char b[])
//'b' is given memory in 'stack' region;'b' has a reference
// to 'a' which in turn refer to 'F'
{
printf("*a is:%c\n",*a);
printf("&a is:%p\n",&a);
//printf("&&a is:%p\n",&(&a));//&()...Inside braces,there must be a
// value..
printf("a's address is:%p\n",a);
printf("a's value is:%s\n",a);
printf("b's address is:%p\n",b);
printf("b's value is:%s\n",b);
printf("*b is:%c\n",*b);
printf("&b is:%p\n",&b);
for(i=0;i<strlen(a);i++)
printf("String as an array of characters is:%c\n",b[i]);
}
Here's The Output:
*a is:F
&a is:00409000
a's address is:00409000
a's value is:F*** Yeahh!
b's address is:00409000
b's value is:F*** Yeahh!
*b is:F
&b is:0028FF20
String as an array of characters is:F
String as an array of characters is:*
String as an array of characters is:*
String as an array of characters is:*
String as an array of characters is:
String as an array of characters is:Y
String as an array of characters is:e
String as an array of characters is:a
String as an array of characters is:h
String as an array of characters is:h
String as an array of characters is:!
