有人可以解释第 3 行发生了什么吗？

Question

vector<long> v = {1,2,3};
long i = v.size();
const long* w = (i != 0) ? &v.front() : NULL;

Can someone explain what is happening on line 3? To me, it feels like v = w. Am I understanding it right?

Answer 1

e1? v1: v2

That's ?: expression where when expression ( e1 ) has true, it returns v1 , else returns v2 .

Here it means point constant pointer (cannot change address once assigned address) to NULL if v has no value inside. Or the first element's address if v has at least one element inside.

Answer 2

it feels like v = w.

No, not at all. v is a std::vector<long> (assuming vector is std::vector ) and w is a const long* . Thats two completely unrelated types. You cannot assign one to the other.

Actually your code is similar to:

vector<long> v = {1,2,3};
const long* w = &v[0];

w is a pointer to the first element in the vector. v.front is just a different way to get a reference to the first element. And because in general we cannot know if v has an element at index 0 , the author added a check:

vector<long> v = {1,2,3};
long i = v.size();
const long* w;
if ( v.size() != 0) w = &v.front();
else                w = nullptr;

And a less verbose way of writing the same is using the conditional operator:

const long* w = ( v.size() != 0 ) ? &v.front() : nullptr;

Depending on the condition, only one side of : is evaluated, hence out-of-bounds access to element 0 in an empty vector can be avoided.

Answer 3

"const long* w = (i?= 0). &v:front(); NULL. " It checks whether vector has elements, If vector has elements. it assigns first elements address to long type pointer? . is ternary operator; example int a=1;int b=2? (a<b): "if true do something"; "if false do something" ;

Answer 4

const long* w = (i?= 0). &v:front(); NULL;

w is a pointer to a long constant - not to be confused with a const pointer to a long . The value of the pointer is being initialized to the result of the ternary expression: (i?= 0). &v:front(); NULL; (i?= 0). &v:front(); NULL;

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The?: operator returns one of two values depending on the result of an expression.

Source: http://www.cplusplus.com/articles/1AUq5Di1/

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If (i != 0) is true then the result will be the address of the reference returned by v.front() (ie: the first element in the vector v ). If the expression evaluates false then w will be a null pointer.

A good rule to follow to ensure you understand any declaration properly is The "Clockwise/Spiral Rule" - understanding this rule can save you some headaches.