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;
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;
Jump to PostBehavior is undefined since ptr2 does not point anywhere.
Jump to PostDid you try compiling it and running it?
s s
Jump to PostThe code you given has some error, are you sure you are giving the right code?
Anyway, the output should be:x x
cause ptr1 has been assigned to the address of var2. When you output the value of *ptr1, it will refer to the value of that address.
What …
Jump to Postvar1 is a double
ptr1 is a pointer to double
...you should be able to handle it from here.
Jump to Postso 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?
Uh, look up a couple posts where I answered that.
Behavior is undefined since ptr2 does not point anywhere.
Sorry i missed a line when i typed it in..could you take another look now..
char var1='s';
char var2='x';
char *ptr1, *ptr2;
ptr1=&var1;
ptr2=&var2;
*ptr2=*ptr1;
cout << *ptr1 << " " << var2 << endl;
Did you try compiling it and running it?
s s
The code you given has some error, are you sure you are giving the right code?
Anyway, the output should be:
x x
cause ptr1 has been assigned to the address of var2. When you output the value of *ptr1, it will refer to the value of that address.
no..because i don't understand how it works..can you please explain it to me...
i need to know the data types of variables such as:
double var1, *ptr1, *ptr2;
float *ptr3;
int var2, *var4....do you know anything about data types?
The code you given has some error, are you sure you are giving the right code?
Anyway, the output should be:x x
cause ptr1 has been assigned to the address of var2. When you output the value of *ptr1, it will refer to the value of that address.
What are you talking about?
i need to know the data types of variables such as:
double var1, *ptr1, *ptr2;
float *ptr3;
int var2, *var4....do you know anything about data types?
???
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
var1 is a double
ptr1 is a pointer to double
...you should be able to handle it from here.
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;
:o sorry, just now i refer to the first code that you given. The right answer is s s.
&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
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
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?
Here's a link:Check it and see, ask if in doubt
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?
Uh, look up a couple posts where I answered that.
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';
-----------------------------------------------------------
output will be
X X
because ptr1 point to the var2 and var2 contain value 'x';
Not necessarily. At that point dereferencing an uninitialized pointer has caused undefined behavior. After that, nothing is guaranteed to work, even if it would otherwise.
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