undefined-behavior

问题 I'm well aware of the difference between class and struct, however I'm struggling to authoritatively say if this is well defined: // declare foo (struct) struct foo; // define foo (class) class foo { }; // instance of foo, claiming to be a struct again! Well defined? struct foo bar; // mixing class and struct like this upsets at least one compiler (names are mangled differently) const foo& test() { return bar; } int main() { test(); return 0; } If this is undefined behaviour can someone point

问题 I've seen the other similar questions and read the defect about it. But I still don't get it. Why is i = ++i + 1 well-defined in C++11 when i = i++ + 1 is not? How does the standard make this well defined? By my working out, I have the following sequenced before graph (where an arrow represents the sequenced before relationship and everything is a value computation unless otherwise specified): i = ++i + 1 ^ | assignment (side effect on i) ^ ^ | | ☆i ++i + 1 || ^ i+=1 | ^ 1 | ★assignment (side

问题 I've seen the other similar questions and read the defect about it. But I still don't get it. Why is i = ++i + 1 well-defined in C++11 when i = i++ + 1 is not? How does the standard make this well defined? By my working out, I have the following sequenced before graph (where an arrow represents the sequenced before relationship and everything is a value computation unless otherwise specified): i = ++i + 1 ^ | assignment (side effect on i) ^ ^ | | ☆i ++i + 1 || ^ i+=1 | ^ 1 | ★assignment (side

问题 I just checked the C++ standard. It seems the following code should NOT be undefined behavior: unsigned int val = 0x0FFFFFFF; unsigned int res = val >> 34; // res should be 0 by C++ standard, // but GCC gives warning and res is 67108863 And from the standard: The value of E1 >> E2 is E1 right-shifted E2 bit positions. If E1 has an unsigned type or if E1 has a signed type and a non-negative value, the value of the result is the integral part of the quotient of E1/2^E2. If E1 has a signed type

问题 I just checked the C++ standard. It seems the following code should NOT be undefined behavior: unsigned int val = 0x0FFFFFFF; unsigned int res = val >> 34; // res should be 0 by C++ standard, // but GCC gives warning and res is 67108863 And from the standard: The value of E1 >> E2 is E1 right-shifted E2 bit positions. If E1 has an unsigned type or if E1 has a signed type and a non-negative value, the value of the result is the integral part of the quotient of E1/2^E2. If E1 has a signed type

问题 I was reading about union in C from K&R, as far as I understood, a single variable in union can hold any one of the several types and if something is stored as one type and extracted as another the result is purely implementation defined. Now please check this code snippet: #include<stdio.h> int main(void) { union a { int i; char ch[2]; }; union a u; u.ch[0] = 3; u.ch[1] = 2; printf("%d %d %d\n", u.ch[0], u.ch[1], u.i); return 0; } Output: 3 2 515 Here I am assigning values in the u.ch but

问题 Suppose I have allocated some memory for storing an int value like this: int *p=new int; Here I created the required memory using new operator and assigned the address of that memory block so that I can access that memory block. Now it's my control to what I store in that memory block. But when I write a statement like this: delete p; we say that I have deleted the dynamically allocated memory. But if I really delete 'd or freed up that memory, should I not be able to access that memory

问题 Suppose I have allocated some memory for storing an int value like this: int *p=new int; Here I created the required memory using new operator and assigned the address of that memory block so that I can access that memory block. Now it's my control to what I store in that memory block. But when I write a statement like this: delete p; we say that I have deleted the dynamically allocated memory. But if I really delete 'd or freed up that memory, should I not be able to access that memory

问题 #include<stdio.h> int main() { void add(); int i=2; add(i++,--i); print("%d",i) } void add(int a,int b) { print("%d %d",a,b); } /*what are a and b's value i am actually not getting the answer why b is 2 */ 回答1: in line 6 where add() is called first args is i++ so it will send value 2 ie value of i to the function and then add 1 now i=3. second args is --i now it will subtract 1 and iwill now be 2 again and then send value 2 to function So I think your answer will print 2 2 (that is value of a

问题 #include<stdio.h> int main() { void add(); int i=2; add(i++,--i); print("%d",i) } void add(int a,int b) { print("%d %d",a,b); } /*what are a and b's value i am actually not getting the answer why b is 2 */ 回答1: in line 6 where add() is called first args is i++ so it will send value 2 ie value of i to the function and then add 1 now i=3. second args is --i now it will subtract 1 and iwill now be 2 again and then send value 2 to function So I think your answer will print 2 2 (that is value of a