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I am writing a short program that displays the argument count (argc), argument vector(argv[]), and environment vector. However, I am unsure how to display the "environment vector" or what it is.
The "environment" parameter, traditionally named envp, is a zero-terminated array of char*.
You can display it like this:
int main(int argc, char* argv[], char* envp[])
{
while (*envp)
{
std::cout << *envp << std::endl;
envp++;
}
}
It's not part of POSIX (or any other standard) but supported by many compilers.
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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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How do I parse a char array such as:
"2019-11-01T17:00:10.000Z"
I want 17:00:10.000 and nothing else. How do I parse with respect to the char array indices?
The simplest and safest way to deal with strings is to use the standard std::string. The std::string has many class member functions that can help with this. One of them is substr that can be used to create a new std::string from a part of a string.
Example:
#include <iostream>
#include <stdexcept>
#include <string>
std::string get_time_str(const std::string& timestamp) {
if(timestamp.size() != 24)
throw std::runtime_error("bad timestamp: "+ timestamp);
return timestamp.substr(11, 12); // start at pos 11 and 12 chars forward
}
int main() {
const char* ts = "2019-11-01T17:00:10.000Z";
std::cout << get_time_str(ts) << '\n';
}
Output:
17:00:10.000
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I came across these two terms
Int (*q)[3][4] and. int q[ ][3][4].
What's the difference these two terms?
And one more question .
Char a[ ]="abcd";
Char *p="abv";
a="ghj";
p="ajk";
Printf("℅s℅s",a,p);
Why this would not compile?
It won't compile because of the line:
a = "ghi"
This assigns a const char* to a char* directly. You can use strcpy to copy the string intead:
strcpy(a, "ghi")
You will still have warnings at this point, because you have not declared p as a const. You can fix this like so:
const char* p = "abv"
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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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If the code is like the below,
void func(std::string str)
{
...
}
void main()
{
std::string p1 = "abcd";
char p2[SOME_LENGTH] = "abcd";
func(p1); // (1)
func(p2); // (2)
}
which way is efficient between (1) and (2)?
They are equally efficient/inefficient. Both involves copying the string and using the copy as the value of the argument 'str'. A better way would be declaring func as
void func(const std::string &str) {
}
This can avoid copying of the string.