passing only first element of the array of chars to a function c++

Question

Basically I have an array of words

 void print(char *str) {
 cout << str <<endl;
 }

 int main() {
 int i =0;
 char* name[] = {"Fred", "John", "Jimmy"};
 print(*name[0]);
 }

I would like to pass only first word to a function, but when I am doing in a way as I am doing right now it passes all the names.

What you have should not compile. – GManNickG Dec 07 '12 at 18:09 — GManNickG, Dec 07 '12 at 18:09

score 4 · Answer 1 · answered Dec 07 '12 at 18:08

4

You have an extra dereference there, the code should be like this:

print(name[0]);

Ideally, your print function should take const char*, because it does not modify the strings passed in.

answered Dec 07 '12 at 18:08

Sergey Kalinichenko

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It should be noted that the code is not quite correct though. It's not const-correct. – Nikos C. Dec 07 '12 at 18:09
Also note that `name` points to string literals, but is not declared as such :) – Nikos C. Dec 07 '12 at 18:11
1

While we're at it, why are we using a C array of C strings, anyway? – chris Dec 07 '12 at 18:29

score 1 · Answer 2 · answered Dec 07 '12 at 18:12

char* name[] is an array of pointers to char, typically strings.

void print(char *str) requires a single pointer to a char as an argument.

By calling print(*name[0]), you are actually taking the first pointer from your name[] array, and dereferencing it to turn it into a single char. Since your function requires a pointer to char, you simply need to call it with print(name[0]) and you'll get the first item in your array.

passing only first element of the array of chars to a function c++

2 Answers2