Printing array of chars using pointer to char.

Question

In main.cpp I must print my char array like this:

const char *str(CValue::TestStringValue0());
cout << ' ' << *str << endl; //must not change

I can't modify this! So I need to print my array of char, but nor first value of array.

TestStringValue look like this..

static const char* TestStringValue0() {...}

Can you show the code for TestStringValue0()? Instead of returning e.g. str you would return str+1 — clcto
– clcto, Commented Nov 21, 2013 at 19:19
What does CValue::TestStringValue0 look like? And what is your question? — Zac Howland
– Zac Howland, Commented Nov 21, 2013 at 19:23
Now I'm having it: static const char* TestStringValue0() '{ return "0,0,0"; }' I must print "0,0,0", not "0" — Myaumyau
– Myaumyau, Commented Nov 21, 2013 at 19:25
Are you asking how to make code that prints a single character magically print a string without changing it? There's no sensible way to do that, and probably no crazy way either. Or have I misunderstood? — Mike Seymour
– Mike Seymour, Commented Nov 21, 2013 at 19:28

FreeNickname · Accepted Answer · 2013-11-21 19:32:51Z

1

I'm not sure I understand you correctly, but if you're not supposed to modify this line:

cout << ' ' << *str << endl; //must not change

Then you can do like this:

const char *tempStr(CValue::TestStringValue0());
const char **str = &tempStr;
cout << ' ' << *str << endl; //must not change

In this case cout will print the whole string returned from TestStringValue0, while in the original code it printed only the first char.

P.S. What a strange condition you have :)

edited Nov 21, 2013 at 19:32

answered Nov 21, 2013 at 19:27

FreeNickname

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2 Comments

andre Over a year ago

Why do you need a temporary variable?

FreeNickname Over a year ago

@andre, since we can't modify the line with count, we need to make *str* a pointer, so we need str` to be a pointer to a pointer. As for the additional variable, I guess, I could avoid it, but I don't feel comfortable with getting address of values just returned from a function :) It's a matter of habit.

Vlad from Moscow · Accepted Answer · 2013-11-21 19:59:10Z

1

const char *str(CValue::TestStringValue0());
const char *p = str;

for ( ; *str; ++str )
{
    cout << ' ' << *str << endl; //must not change
}

:)

edited Nov 21, 2013 at 19:59

answered Nov 21, 2013 at 19:35

Vlad from Moscow

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