How to correctly pass a String array as function argument?

Question

I know this has been asked numerous times before. However, I'm unable to get rid of a warning.

void function (char** cppStringArray);
int main(void) {
    char cStringArray[5][512]={"","","","",""}; //Array of 5 Strings (char arrays) of 512 characters
    function (cStringArray); //warning: incompatible pointer type
    return 0;
}

How do I get rid of the warning? It works if I declare the Stringarray as char* cStringArray[5].

You should not pass modifiable string without passing size information. Either use const or supply size — stefan bachert
– stefan bachert, Commented Feb 20, 2014 at 11:55
@stefanbachert I know I am supplying the size just did not include it in my post. — H_squared
– H_squared, Commented Feb 20, 2014 at 11:56

Sergey L. · Accepted Answer · 2014-02-20 11:50:06Z

1

If your string array will remain with those sizes then the best way is to use

void function (char (* cppStringArray)[512], size_t num_strings);

pass as

function(cStringArray, sizeof(*cStringArray)/sizeof(cStringArray));

Problem is that char** is not equivalent to char (*)[512]. The former is a pointer to pointer to char. The latter is a pointer to a block of 512 characters.

answered Feb 20, 2014 at 11:50

Sergey L.

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2 Comments

H_squared Over a year ago

Perfect, thank you. Apparently the parentheses around *cppStringArray are important. I was trying that without the parentheses!

H_squared Over a year ago

@ajay i was talking about "parentheses around *cppStringArray " i.e. void function (char (* cppStringArray)[512], size_t num_strings); and not void function (char *cppStringArray[512], size_t num_strings);

quetzalcoatl · Accepted Answer · 2014-02-25 10:03:24Z

1

Define the function like below.

void function (char cppStringArray[5][512]);

edited Feb 25, 2014 at 10:03

quetzalcoatl

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answered Feb 20, 2014 at 11:59

Amit

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