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using namespace std;
int main()
{
int n, *p1, *p2;
n = 10;
p1 = &n;
p2 = p1;
(*p1)++;
(*p2)++;
cout << *p1 << " " << *p2 << " "<< n << endl ;
return 0;
}
*p1 refers to the value pointed by the pointer p1. (*p1)++ will increment the value of n by 1 and (*p2)++ will again do the increment on n, since it is pointing to the same location of p1. So the n will be incremented to 12. *p1,*p2 and n will thus have 12. So it prints 12.
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int* p= NULL;
int** pp = &p;
cout << &p << endl;
cout <<*pp << endl;
I think the output results of the two should be consistent, but the first output is the address of p, and the second is 00000000. I want to know why.
You are mistaken, pp is the address of p
int** pp = &p; // set pp to the address of p
But *pp is the value of whatever pp is pointing at. In this case that is p which has a value of NULL.
The code is essentially no different to this
int x = 123;
int* px = &x;
cout << &x << endl; // print address of x
cout << *px << endl; // print value of x i.e. 123
The only difference being the type of the original value (int vs int*).
pp holds the address of p, so print pp, not *pp.
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int main() { vector g1; vector :: iterator i; vector :: reverse_iterator ir;
for (int i = 1; i <= 5; i++)
g1.push_back(i);
cout << "Output of begin and end\t:\t";
for (i = g1.begin(); i != g1.end(); ++i)
cout << *i << '\t';
cout << endl << endl;
cout << "Output of rbegin and rend\t:\t";
for (ir = g1.rbegin(); ir != g1.rend(); ++ir)
cout << '\t' << *ir;
return 0;
}
Here in this code variable "i" has been declared as a iterator as well as a variable inside a for loop. isn't that a error?
If we see the first for loop it say that the loop will run till i!=g1.end() that means that the value of *(g1.end()) should not be displayed by *i but it is giving. ide shows output 1 2 3 4 5 for me it should be 1 2 3 4.
i is defined as an iterator in the argument list. When you redefine it in the first for loop, this is a new definition only for the scope of the loop -- this is perfectly legal, though not good practice.
vector::end() points to memory after the final item, not to the final item. So, yes, the final item in the vector will be printed.
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I am very new at coding and I just had a quick question.
How would you find the nth character from a string in an array for c++?
so for example
{"Hey you" "echo" "lake" "lend" "degree"}, n=0
would equal "hello"
thanks! I'm not going to ask for the full code but just tips on where to get started.
Here are some examples:
unsigned int n = 3;
static const char hello[] = "hello";
cout << hello[n] << "\n";
const std::string apple = "apple";
cout << apple[n] << "\n";
static const char * many[] = {"These", "are", "many", "strings"};
cout << many[n][n] << "\n";
cout << many[n] << "\n";
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Fill up the blanks with appropriate keyword to get the desired output according to the test cases. (Language use c++)
Sample Test Cases
input output
Test Case 1 4 square = 16, ++ square = 25
test case 2 -8 square =64, ++ square = 49
#include <iostream>
using namespace std;
______ int SQUARE(int x) { ______ x * x; }
int main() {
int a , b, c;
cin >> a ;
b = SQUARE(a);
cout << "Square = " << b << ", ";
c = SQUARE(++a);
cout << "++ Square = " << c ;
return 0;
}
Second blank place is "return"
First may be "inline" or nothing
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why is casting working in the first case but not in second cout after masking one of the characters in char*
#include <iostream>
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
using namespace std;
int main()
{
int n = 10;
char* ch = (char*) calloc(6, sizeof(*ch));
ch = strdup("ab");
cout << strlen(ch) << endl;
int* p = (int*) ch;
cout << (char*)p << endl;// works fine it prints "ab"
*p = *p & 65280;
cout << "cast not working\t" << (char*)p << endl; // it does not work here
free(ch);
return 0;
}
Written as hex, 65280 is 0x0000FF00. So, on common systems with int being 4 bytes, you have set ch[0] to be 0. This is a null terminator, so when you try to print the string, you see an empty string.
Note: writing to *p causes undefined behaviour by writing past the end of the allocated area too; strdup("ab") allocates 3 bytes. On common systems this probably will have no ill effect as heap allocations are done in chunks of a certain size.