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    <p>AFAIK this is what you intend:</p> <pre><code>struct stu { char **argv; // *argv[] will give you "error: flexible array member not at end of struct" int a; }; struct stu Stu; // Allocate Stu as a global stu struct fun(int argc, char * argv[]) {} int main() { char input[128]; char *argies[2]; // Static allocation of space for argv to point to struct stu *ptr = &amp;Stu; ptr-&gt;argv = argies; // You need to assign argv to point to allocated space ptr-&gt;argv[0] = "name"; ptr-&gt;argv[1] = input; fun(2, ptr-&gt;argv); // is 2nd argument we passing member of array argv?? } </code></pre> <p>If you do not declare an actual stu struct, then ptr wiill be initialized to <a href="https://stackoverflow.com/questions/1597405/what-happens-to-a-declared-uninitialized-variable-in-c-does-it-have-a-value">zero</a>. Also if you do declare a static struct stu (Stu above) but do not initialize the argv pointer, the argv pointer will be initialized to zero. Not pointing to "who knows where" but pointing to a very specific place.</p> <p>Note that input is not NULL terminated, in fact it is not initialized, so the variable name "argv" might be misleading because most programmers might assume that any char **argv holds NULL terminated C strings. Also, you probably want to dimension argies to 3 and put a NULL in the third position.</p> <p>You can pass ptr->argv[0] (the string "name") as *(ptr->argv), but the fun prototype would have to be fun(int argc, char * argv).</p> <p>**ptr->argv is a char.</p> <p>*ptr is a struct stu.</p> <p>**ptr is an error.</p>
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    1. POCan we pass structure member this way?
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    1. USCommunity
    1. USJonathan Ben-Avraham
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    1. COThanks Jonathan ,what i wanted to know is ,here fun(2,(**)ptr) in second arugment are we passing member of array input??
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      1. USAmit Singh Tomar
 

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