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Need help figuring out C++ Pointers
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ok i ran this code now and i get s s not x x...are you sure x x is right????
char var1='s';
char var2='x';
char *ptr1, *ptr2;
ptr1=&var1;
ptr2=&var2;
*ptr2=*ptr1;
cout << *ptr1 << " " << var2 << endl;
char var1='s';
char var2='x';
char *ptr1, *ptr2;
ptr1=&var1;
ptr2=&var2;
*ptr2=*ptr1;
cout << *ptr1 << " " << var2 << endl;
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Join Date: May 2004
Posts: 256
Reputation: 
Solved Threads: 6
Posting Whiz in Training
&val means address of the varialble var
*p (if is a pointer (int *p
) means value at address pointed at by p.Changes this will change the variable it points to.
int val = 1;
p=&val;
val+=10
cout<<val<<" *p="<<*p; //output is the same for both
(*p) += 4; //changes the value of val
cout<<val<<" *p="<<*p; //output is the same for both
The outputs are:
11 11
15 15
*p (if is a pointer (int *p
) means value at address pointed at by p.Changes this will change the variable it points to.int val = 1;
p=&val;
val+=10
cout<<val<<" *p="<<*p; //output is the same for both
(*p) += 4; //changes the value of val
cout<<val<<" *p="<<*p; //output is the same for both
The outputs are:
11 11
15 15
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Originally Posted by hfick
the next question asks says i am supposed to state the data type of each variable...
double var1, *ptr1, *ptr2;
float *ptr3;
int var2, *var4
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Join Date: May 2004
Posts: 256
Reputation:
Solved Threads: 6
Posting Whiz in Training
Here's a link:Check it and see, ask if in doubt
http://www.daniweb.com/techtalkforums/thread9878.html
http://www.daniweb.com/techtalkforums/thread9878.html
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Join Date: Apr 2004
Posts: 4,414
Reputation: 
Solved Threads: 248
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Originally Posted by hfick
so what would the data type of ptr1 be? would it be just ptr1 since it doesn't have the (*) in front of it or how does that work?
"One of the methods used by statists to destroy capitalism consists in establishing controls that tie a given industry hand and foot, making it unable to solve its problems, then declaring that freedom has failed and stronger controls are necessary." --Ayn Rand
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Originally Posted by hfick
I am supposed to figure out what the output produced would be from this equation..can anyone help me and explain it to me
char var1='s';
char var2='x';
char *ptr1, *ptr2;
ptr1=&var2;
*ptr2=*ptr1;
cout << *ptr1 << " " < var2 << endl;
output will be
X X
because ptr1 point to the var2 and var2 contain value 'x';
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Originally Posted by manoj9_5
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output will be
X X
because ptr1 point to the var2 and var2 contain value 'x';
I'm here to prove you wrong.
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