C++ Pointer Char - c++

Question

I'm new to C++ and I'm now learning about pointers. I'm trying to understand this program:
#include <iostream>
int main() {
char *text = "hello world";
int i = 0;
while (*text) {
i++;
text++;
}
printf("Char num of <%s> = %d", text, i);
}
It outputs:
Char num of <> = 11
But why not this:
Char num of <hello world> = 11

Answer 1

As your increase text until it points to '\0',
the string printed is empty.

Answer 2

You changed the value of text before printing it out:
#include <iostream>
int main() {
// text starts pointing at the beginnin of "hello world"
char *text = "hello world"; // this should be const char*
int i = 0;
while (*text) {
i++;
// text moves forward one character each time in the loop
text++;
}
// now text is pointing to the end of the "hello world" text so nothing to print
printf("Char num of <%s> = %d", text, i);
}
The value of a char pointer (like text) is the location of the character that it is pointing to. If you increase text by one (using text++) then it points to the location of the next character.
Not sure why you #include <iostream> but use printf(). Typically in C++ code you would do this:
std::cout << "Char num of <" << text << "> = " << i << '\n';
A working example of what you are trying to do:
int main()
{
const char* text = "hello world"; // should be const char*
int length = 0;
for(const char* p = text; *p; ++p)
{
++length;
}
std::cout << "Char num of <" << text << "> = " << length << '\n';
}

Answer 3

First of all - don't use char-string in c++! Use std::string.
Your while loop continues until it reach zero which is the string termination, so %s is just an empty string. The '<' and '>' is still printed even if the string is empty.
Your text pointer start as the following chars:
'h','e','l','l','o',' ','w','o','r','l','d','\0'
After the first loop, text points to:
'e','l','l','o',' ','w','o','r','l','d','\0'
After second loop, text points to:
'l','l','o',' ','w','o','r','l','d','\0'
And so on.
The while-loop continues until text points to '\0' which is just an empty string, i.e. "".
Consequently %s doesn't print anything.
In c++ do:
int main() {
std::string text = "hello world";
cout << "Char num of <" << text << "> = " << text.size() << std::endl;
}

Answer 4

text is a pointer to a char. So when you increment the pointer you make it to point to the next element. The end of a string in C is delimited by the null character. So you are incrementing the pointer until it points to the null character. After that, when you call printf it doesn't print anything because it prints characters until reaches the null character but it is already at the null character.
This could be a quick fix:
#include <stdio.h>
int main() {
char *text = "hello world";
int i = 0;
while (*text) {
i++;
text++;
}
printf("Char num of <%s> = %d", text-i, i);
}

Answer 5

After the loop pointer text points to the terminating zero of the string literal because it is increased in the loop.
Also function printf is declared in header <cstdio> And string literals in C++ have types of constant arrays (Though in C they have types of non-constant arrays. Nevertheless in the both languages you may not change string literals)
You can rewrite the program either like
#include <cstdio>
int main() {
const char *text = "hello world";
int i = 0;
while ( text[i] ) ++i;
std::printf("Char num of <%s> = %d\n", text, i);
}
Or like
#include <cstdio>
int main() {
const char *text = "hello world";
int i = 0;
for ( const char *p = text; *p; ++p ) ++i;
std::printf("Char num of <%s> = %d\n", text, i);
}
Also it is better instead of C function printf to use standard C++ stream operators.
For example
#include <iostream>
int main() {
const char *text = "hello world";
size_t i = 0;
for ( const char *p = text; *p; ++p ) ++i;
std::cout << "Char num of <" << text << "> = " << i << std::endl;
}