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I'm trying to make a login/register project and I have difficulties in declaring the char* tempUsername from this code (SIGSEVG segmentation fault)
char *tempUsername, *tempPassword, *tempPasswordConfirm, *tempSecurityQuestion;
/*
no other declaration for tempUsername here
*/
std::cout<<"Enter your new username:\n";
std::cin>>tempUsername;
//process stops here
if(fileSearch(newFilename(tempUsername))) {
std::cout<<"Username already exists! Choose another username!\n";
}
else {
std::cout<<"Enter your password:\n";
std::cin>>tempPassword;
std::cout<<"Confirm your password:\n";
I'm having a hard time understanding anything about pointers, so any advice is more than helpful!
char *tempUsername
std::cin>>tempUsername;
The problem here is that your pointer is uninitialised. When you try to extract from the input stream into the uninitialised pointer, the behaviour of the program will be undefined. Don't do this.
Your goal seems to be to read a string of user input. A solution that I can recommend is to use the std::string class:
std::string tempUsername;
std::cin >> tempUsername;
No need to use a pointer to achieve this.
I can use a char* as an array of chars, is that true?
It is not true in general. If you have a char* that points to an element of an array of char, then you can use that char* as an iterator to access the elements of the array.
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How can make a program typing with c++?
For example with (char) can type a grapheme but i want type word.
char x;
x=='c';
But i want type for example 'equation'.
Someone know how can i do this?
Use a std::string. You can also use char* but i recommend you the first one. Remember that string goes within "" instead of ''.
string x; //You have to define what is 'x'.
x=="equation";
If you use char*, it behaves as an array of chars.
#include <string> // paste this line on top of the file
std::string name = "Bob"; // Use double quotes here.
if(name == "Bob")
{
std::cout << name << std::endl; // works as you expected
}
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I am trying to modify a c string in C++.
void modify(char* s)
{
s[0] = 'a';
}
If I do this, there will be some undefined behavior and can't run.
Let's assume s[0] is valid. I know char* s is immutable. Is there any possibility that I can modify the s[0] in place, which means, without creating a new string. Do the modification on the original string.
I think you might be misunderstanding some other answers you have seen on the web.
It is only undefined behavior to modify a string constant, not any char*.
As long as you strdup the constant string into a non-constant string, you can make whatever changes you want with it, because it is now in a mutable region of memory.
#include <stdio.h>
#include <string.h>
void modString(char* changeMe) {
changeMe[0] = 'g';
}
int main(){
char* foo = strdup("food");
puts(foo);
modString(foo);
puts(foo);
free(foo);
}
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Let's take the following code for example
int number = 1;
char * charsequence = (char *)&number //casting the address of number to char *
std::cout << charsequence << endl;
The above snippet produces the following result:
☺
If I change the number then a different symbol appears.
That's fine, the real question is why do I always get the same symbols even though the memory location (&number) is different on each run? And the most important part is why do I get symbols instead of my memory address?
I assume that the casting I did is not working as I thought it was.
Edit: Since it's closed as unclear of what I'm asking, here's there real question:
How do i print the memory address of an object to the console?
At the time of this edit, this question was already answered. See the accepted answer
std::ostream has a special overload for operator<< which, when you give it a char *, will not print the value of the pointer, but instead assume that you gave it a pointer to the first element in an array of characters that is null-terminated, and then try to print the entire array up to the terminator. (This overload allows you to print C strings in a sort-of natural looking syntax.)
Since your pointer does not in fact point to the first element of a null-terminated array, your program has undefined behaviour.
If you want to print the pointer value, you should use a void pointer:
std::cout << static_cast<void*>(&number) << "\n";
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I am trying initialize element of array
array[m][n] == char("X");
after printing that element i'm getting the value of it equals д (Russian d); how deal with it, and I'm not even able to initialize that element without parsing const char to char.
You have to write simply as
array[m][n] = 'X';
where 'X' is a character literal.
Or if you like very much string literals then:)
array[m][n] = *"X";
or
array[m][n] = "X"[0];
EDIT: I am sorry. You have also to use the assignmnet operator (=) instead of the comparison operator (==)
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For example:
string str[3];
void foo(char** str)
{
//do something to str...
}
How to pass str[] to function foo in a convenient way?
The function expects an array of pointers, so you'll have to make one from your array of strings:
std::vector<char*> pointers;
for (auto & s : str) {
pointers.push_back(&s[0]);
}
foo(&pointers[0]);
Beware that this may not be valid if the function modifies the pointers, or the strings they point to. A better option would be to avoid mixing C and C++ style string handling, if possible.