Answer / avinash

In this question ,two wrong thins ----
1.this is an array of char pointer so use
strcy(text[no.],"Hello World");

2.
we are copying a string without allocating memory to pointer . This is bug code .

correct solution :----

char *string()
{
char *text[20];
text[0]=malloc(20*sizeof (char));
strcpy(text,"Hello world");
return text;
}

Answer / vignesh1988i

as for as i know , there is only one error..... you have
declared text as array of pointers and not as character data
array..... so this text can only accept addresses.... :)

char *text[20] means you are going to store 20 addresses in
this array..... When you store addresses using arrays , the
that is called array of pointers....

if u declare : char text[20] , this will work correctly..



thank u

Answer / qint

1. returning address of a local variable.
2. wrong parameter passed to strcpy()

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