This question already has answers here:
How do I use arrays in C++?
(5 answers)
Closed 7 years ago.
I saw the following line of code in C++. I have trouble understanding it. I hope that I can get some help here.
// Example program
#include <iostream>
#include <string>
using namespace std;
int main()
{
int a[] = {1,2,3,4};
cout<<*a+1<<;
cout<<a[1];
}
Question:
I don't understand how *a+1 works. It seems pretty unintuitive - are we adding 1 to the array here?
To understand the expression *a+1 you should consider the operator priority:
the dereference * has higher priority, so a (the name of an array is also the pointer to the first element) is dereferenced, giving the value 1
Then, 1 will be added to the value.
operator priorities are describe here http://en.cppreference.com/w/cpp/language/operator_precedence
In the expression *a + 1, the array a is evaluated in pointer context, so it points to the first value in the array.
Next, *a dereferences that pointer (meaning it gets the value the pointer points to), and that gives you the first element in the array which is 1.
Finally, 1 is added to that value and printed, so it outputs 2.
First of all this statement
cout<<*a+1<<;
^^^^
is syntactically wrong.
I think you mean something like
cout << *a+1 << endl;
Expression *a is equivalent to a[0]. So you could write instead
cout << a[0] + 1 << endl;
Take into account that semantically a[0] + 1 is not the same as a[1] though applied to your example the both expressions will yield the same result.:)
Related
This question already has answers here:
Accessing an array out of bounds gives no error, why?
(18 answers)
Closed 4 years ago.
int main()
{
int n[] = {2};
cout <<n[1] << endl;
return 0;
}
In this code, I get some numbers,also for n[2],n[3] etc. I got ë for a string array and # for a char array. I don't think it is the memory location so what is this?
int n[] = {2};
Here you are declaring an array with one element. So now n[0] is 2. That's the only element, so everything else is out of bounds. Therefore this here:
cout <<n[1] << endl;
Is undefined behavior, because you're accessing the second element, which doesn't exist. Remember, array indexing starts at 0, so element 0 is the first one and 1 is the second one.
So why does it print a rubbish value, then? Since it's undefined behavior, there's no guarantee for what it does, but what's most likely going to happen is that it's calculating the respective memory location and prints the data at the resulting address, interpreted as int. That usually results in some rubbish value.
This question already has answers here:
Adding an offset to a pointer
(3 answers)
Closed 5 years ago.
In the cplusplus page for the inner_product module they gave a code example:
int init = 100;
int series1[] = {10,20,30};
int series2[] = {1,2,3};
std::cout << "using default inner_product: ";
std::cout << std::inner_product(series1,series1+3,series2,init);
std::cout << '\n';
Where I saw them use series1+3 when calling the inner_product funtcion.
What exactly does adding "3" to the array variable do?
What exactly does adding "3" to the array variable do?
The array operand decays into a pointer to first element of the array. 3 is added to that pointer, so that the result is a pointer to 3 elements after the first (i.e. the element at index 3), which is immediately outside the bounds of the array. The addition is equivalent to std::end(series1) which would be more idiomatic in my opinion.
I'm assuming that series1 on itself is a pointer to the first element of the array
You assume wrong. series1 is not a pointer, but an array. However, an id-expression (standardese for the name of the variable) of an array will decay into a pointer in value contexts.
This question already has answers here:
Post-increment and Pre-increment concept?
(14 answers)
Closed 7 years ago.
just today I stumbled across this short piece of code that confused me a little.
#include <iostream>
#include <iterator>
int main()
{
int array[] = {0,1,2,3,4,5,6,7,8,9};
auto start = std::begin(array);
while (start != std::end(array))
std::cout << *start++ << std::endl;
}
The thing that is confusing me here is that the 0 is the first output. I read a lot of posts regarding the order of the 2 operators and every single one said: "start" would be incremented first, THEN dereferenced. But std::begin() returns an iterator to the beginning of the array. With this being said, if I increment the pointer to the beginning of the array first before dereferencing it, shouldn't my first output be the 1?
Thanks in advance!
*start++ uses the post increment operator. With post increment the item is incremented but the value returned is the value before incrementing. You are dereferencing that value so that is why the output starts at 0.
This question already has answers here:
Why does i[arr] work as well as arr[i] in C with larger data types?
(5 answers)
Closed 7 years ago.
The following code
#include<stdio.h>
int main()
{
int arr[] = {10,20,30};
cout << -2[arr];
return 0;
}
prints -30. How? Why?
In your case,
cout<<-2[arr];
gets translated as
cout<<-(arr[2]);
because,
array indexing boils down to pointer arithmatic, so, the position of array name and the index value can be interchanged in notation.
The linked answer is in C, but valid for C++ also.
regarding the explicit ()s, you can check about the operator precedence here.
Look at this statement
cout << -2[arr];
First, know that even though it looks strange the following is true
2[arr] == arr[2]
That being said operator[] has higher precedence than -. So you are actually trying to invoke
-(arr[2])
In C and C++, 2[arr] is actually the same thing as arr[2].
Due to operator precedence, -2[arr] as parsed as -(2[arr]). This means that the entire expression evaluates to negating the 3rd element of arr, i.e. -30.
This question already has answers here:
pointer increment and dereference (lvalue required error)
(4 answers)
Closed 9 years ago.
int main()
{
int[] x={1,2,3,4,5};
printf("%d",x);
printf("%d",*x);
printf("%d",++*x);
printf("%d",*x++); //Here Lvalue required erroe is genrerated
}
Can someone please explain me what is meaning of this error and why it is generating here
You can't increment an array.
*x++ is the same as *(x++). Maybe you wanted (*x)++ instead?
The correct way to declare array in c/c++ is :
int x[]={1,2,3,4,5};
The error is because you can not use ++ on array
In this statement:
printf("%d",*x++);
You are applying the ++ operator to the array itself, which is not possible. This gives you the lValue error.
Alternatively, in this one:
printf("%d",++*x);
You are applying the ++ operator to the value pointed to by the array (its first element), which of course works fine and just prints the value of next element.
Try
printf("%d",(*x)++);
You can't increment an array. Try this:
int main()
{
int x[]={1,2,3,4,5};
printf("%d",x);
printf("%d",*x);
printf("%d",++*x);
int *y=x;
printf("%d",*y++); //you can increment a pointer though
}
its ok the increment in array makes no sense.But array is just a pointer variable. I mean if I print the value of x in my program(which is an array) it will print address of a[0].It mean it is pointer to a[0].So when I can use increment on a simple pointer variable then why I can not use increment operation on array pointer.