Can someone please explain why the output of the below code snippet is 20 ?
int i = 10;
cout << i << endl;
int &r = i;
r = 20;
cout << i << endl;
First, variable i stores the integer value 10
Then 10 is displayed.
Then the address of r (memory location of r) is set to i which is 10 And then r becomes 20 But why i changes to 20 as well?
The integer content of r has changed, not address of it (memory location of it).
Thanks,
The variable r is a refernce to i, it's like a pointer except that instead of saying *r = 20;
you just say r = 20; and that changes the value of r.
When you make a reference it's almost like a pointer so when you can changed r you actually changed what r was pointing to which is also i.
Actually to be more precise the reference is not a pointer or an address, it is the object. At least in c++ world.
Think about it as:
int i = 10;
cout << i << endl;
int *p = &i;
*p = 20;
cout << i << endl;
This is basically what's happening behind the scenes
Related
The code successfully compiles it but I can't understand why, for certain values of number, the program crashes and for other values it doesn't. Could someone explain the behavior of adding a long int with a char* that the compiler uses?
#include <iostream>
int main()
{
long int number=255;
std::cout<< "Value 1 : " << std::flush << ("" + number) << std::flush << std::endl;
number=15155;
std::cout<< "Value 2 : " << std::flush << ("" + number) << std::flush << std::endl;
return 0;
}
Test results:
Value 1 : >
Value 2 : Segmentation fault
Note: I'm not looking for a solution on how to add a string with a number.
In C++, "" is a const char[1] array, which decays into a const char* pointer to the first element of the array (in this case, the string literal's '\0' nul terminator).
Adding an integer to a pointer performs pointer arithmetic, which will advance the memory address in the pointer by the specified number of elements of the type the pointer is declared as (in this case, char).
So, in your example, ... << ("" + number) << ... is equivalent to ... << &""[number] << ..., or more generically:
const char *ptr = &""[0];
ptr = reinterpret_cast<const char*>(
reinterpret_cast<const uintptr_t>(ptr)
+ (number * sizeof(char))
);
... << ptr << ...
Which means you are going out of bounds of the array when number is any value other than 0, thus your code has undefined behavior and anything could happen when operator<< tries to dereference the invalid pointer you give it.
Unlike in many scripting languages, ("" + number) is not the correct way to convert an integer to a string in C++. You need to use an explicit conversion function instead, such as std::to_string(), eg:
#include <iostream>
#include <string>
int main()
{
long int number = 255;
std::cout << "Value 1 : " << std::flush << std::to_string(number) << std::flush << std::endl;
number = 15155;
std::cout << "Value 2 : " << std::flush << std::to_string(number) << std::flush << std::endl;
return 0;
}
Or, you can simply let std::ostream::operator<< handle that conversion for you, eg:
#include <iostream>
int main()
{
long int number = 255;
std::cout<< "Value 1 : " << std::flush << number << std::flush << std::endl;
number = 15155;
std::cout<< "Value 2 : " << std::flush << number << std::flush << std::endl;
return 0;
}
Pointer arithmetic is the culprit.
A const char* is accepted by operator<<, but will not point to a valid memory address in your example.
If you switch on -Wall, you will see a compiler warning about that:
main.cpp: In function 'int main()':
main.cpp:6:59: warning: array subscript 255 is outside array bounds of 'const char [1]' [-Warray-bounds]
6 | std::cout<< "Value 1 : " << std::flush << ("" + number) << std::flush << std::endl;
| ^
main.cpp:8:59: warning: array subscript 15155 is outside array bounds of 'const char [1]' [-Warray-bounds]
8 | std::cout<< "Value 2 : " << std::flush << ("" + number) << std::flush << std::endl;
| ^
Value 1 : q
Live Demo
In this Book object I created a system to keep track of the time a book has been checked out... there is a tm* called dateCheckedOut to store the date that the book is checked out.
int Book::getHeldTime()
{
time_t now ;
time(&now);
tm* t = localtime(&now);
double difference = difftime(now, mktime(dateCheckedOut))/(60 * 60 * 24);
cout << dateCheckedOut->tm_mon << dateCheckedOut->tm_mday << dateCheckedOut->tm_year << endl; //prints out 231117, which is correct.
cout << t->tm_mon << t->tm_mday << t->tm_year << endl; //prints out 34117, which is also correct
cout << difftime(now, mktime(dateCheckedOut)) << endl; //prints out 0
cout << difference << endl; //prints out 0;
return (int)(difference); //returns 0
}
I am quite confused because I checked the date when its checked out and the date when it is loaded, both are correct, but the difftime function just returns 0. Is there anything that might cause this code to not work? Thank in advance!
P.S. dateCheckedOut only has tm_mday, tm_mon, and tm_year set to the correct value, the rest are all not set. Is that a problem?
I'm working on a mini-compiler and I reached the code generation level.
I want to store a string dynamically to the heap segment, so I wrote this C++ code (cg is the file I'm generating assembly code to) :
int length = strlen("abc");
cg << "\tori\t$a0,$0," << (length + 3) * 4 << endl; // reserve space for type + length + null + size
cg << "\tori\t$v0,$0,9" << endl;
cg << "\tsyscall" << endl;
increamentSP();
cg << "\tsw\t$v0,0($sp)" << endl;
cg << "\tori\t$t1,$0,1" << endl; // store the type
cg << "\tsw\t$t1,0($v0)" << endl;
cg << "\tori\t$t1,$0," << length << endl; // store the length
cg << "\tsw\t$t1," << 4 << "($v0)" << endl;
for (int i = 0; i < length; i++)
{
cg << "\tori\t$t1,$0," << (int)p->val[i] << endl; // store the char
cg << "\tsw\t$t1," << (2 + i) * 4 << "($v0)" << endl;
}
cg << "\tsw\t$0," << (2 + length) * 4 << "($v0)" << endl;
Basically, what I'm trying to do is :
First, I want to store a flag (1) referring that this type is a string to the first location.
Then, I want to store the size of my string to the second location.
After that, I want to store my strings chars to the next locations.
Finally, I want to store a null-terminating char to the last location.
My problem is that when I try to print my string using a code like this :
la $a0,8($t0)
ori $v0,$0,4
syscall
The result is that mips prints only the letter 'a', how can I print a string stored like this? or is there any better way to store my string?
p.s. I know I can use .asciiz in the .data segment, but the problem is that in my code I might edit the string, so I don't exactly know what my string would become.
Can any one help me with that?
When I execute this, every time I input 7 the PC makes a "beep". Can someone explain me why? I use this header: http://www.stroustrup.com/Programming/std_lib_facilities.h
int main()
{
double d = 0;
while (cin >> d){
int i = d;
char c = i;
int i2 = c;
cout << "d==" << d
<< " i==" << i
<< " i2==" << i2
<< " char==(" << c << ")\n";
}
}
ASCII character 7 is the bell character dating back to the ancient days of teletype...
#include <iostream>
using namespace std;
int main ()
{
int size = 0;
int* myArray = new int [size + 1];
cout << "Enter the exponent of the first term: ";
cin >> size;
cout << endl;
for (int i = size; i >= 0; --i)
{
cout << "Enter the coefficient of the term with exponent "
<< i << ": ";
cin >> myArray[i];
}
for (int i = size; i >= 0; --i)
{
cout << i << endl;
}
return 0;
}
Why am I getting an assertion error on input greater than 2? This is the precursor to a polynomial program where the subscript of the array is the power of each term and the element at array[subscript] is the coefficient.
Your array is allocated to be an int[1]. It needs to be allocated after you read in the size value.
You are initializing your array when size = 0, giving an array size of 1
You get your assertion error when you go outside of the array bounds (1).
myArray always has size 0 + 1 = 1. i starts out at whatever the user inputted, and the first array access you make is myArray[i]. So, say the user inputs 5, your array has size 1 and you access myArray[5]. It will fail!
I would allocate the array AFTER you input size.