Question

Can someone explain to me why buff doesn't have the data that i put inside of the char pointer name "data"

29 int main) 30 putsC"Hello World; int yes 2; int finally -1; char *buff; int ret = 0; char *hh buff malloc(7) printf("buff %p \n", &buff); ret = puff(yes, buff, finally) ; 32 36 37 39 40 41 42 . 43 putsC"" putsC"buff"; for(int i-0; i < 6; i++) printf("%C·buff[i]); putsC"done"); 45 46 47 return 0 input Hello World buff 0x7fff79b83f08 data 0x7fff79b83ec8 data 20x7fff79b83ec8 c dej a z data cdej a z buff done

Answer 1

There are certain errors with the code, first is line 37 where pointer conversion is not done right, you need to use the pointer conversion i.e. buffer=(char *)malloc(7); and also use an exception in case the memory is not allocated i.e. if(buffer!=NULL)  

Moreover, I couldn't understand the the use of function puff, where parameter y is never used.

Take care of these errors and the code shall work fine.

Regards