What will print out? main() { char *p1=?name?; char *p2; p2=(char*)malloc(20); memset (p2, 0, 20); while(*p2++ = *p1++); printf(?%sn?,p2); }

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Vinod Kulkarni

Freelance technical trainer / consultant (Embedded systems, Automation / Scripting) : C, Linux, Embedded systems, python, perl, tcl

Copying will happen but pointer will increment and point at the end, so will not print anything as you have already hit at the end.
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p2 = empty string print statement will output = n
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int main() { char *p1="name"; printf("%s\n",p1); char *p2; p2=(char*)malloc(20); // memory allocate size 20 byte,initialize with garbage value that,s why you use memset to clear memory memset (p2, 0, 20);...
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int main() { char *p1="name"; printf("%s\n",p1); char *p2; p2=(char*)malloc(20); // memory allocate size 20 byte,initialize with garbage value that,s why you use memset to clear memory memset (p2, 0, 20); //use memset to clear memory while(*p2++=*p1++) //when pointer P2 point to NULL while loop is terminated { printf("%s\n",p2); //NULL value print } } ans is NULL read less
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C++, C, Embedded, Linux Expert

Out put of this will be n because in while loop you are assigning the value p2 from p1 and incriminating the both pointers p1 and p2 due to this your pointers p1 and p2 will point to the null. So when you try to pointer p2 it will point to nothing and it will print only n.
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