Jeong-taek  Lee Jeong-taek Lee - 22 days ago 8
C++ Question

how to use vector<T>::reverse_iterator with one element

I used poll() with std::vector.
registed listen socket.

std::vector<struct pollfd> fds;
fds.push_back(server_sock);


and add new client socket or connected client session do something.

// poll() ...
for(std::vector<struct pollfd>::reverse_iterator it = fds.rbegin(); it != fds.rend(); it++) {
if (it->fd == server_sock) {
struct pollfd newFd;
newFd.fd = newClient;
newFd.events = POLLIN;
fds.push_back(newFd);
} else {
// do something.
}
}


but the reverse_iterator does not work properly when there is a 1 or 2 or 4 vector's element. I don't understand why this work.

attached sample code.

typedef struct tt_a {
int a;
short b;
short c;
} t_a;

vector<t_a> vec;
for (int i = 0; i < 1; i++) {
t_a t;
t.a = i;
t.b = i;
t.c = i;
vec.push_back(t);
}

for(vector<t_a>::reverse_iterator it = vec.rbegin(); it != vec.rend(); it++) {
if (it->a == 0) {
t_a t;
t.a = 13;
t.b = 13;
t.c = 13;
vec.push_back(t);
}

printf("[&(*it):0x%08X][it->a:%d][&(*vec.rend()):0x%08X]\n",
&(*it), it->a, &(*vec.rend()));
}

printf("---------------------------------------------\n");

for(vector<t_a>::reverse_iterator it = vec.rbegin(); it != vec.rend(); ++it) {
if (it->a == 3) {
it->a = 33;
it->b = 33;
it->c = 33;
}
printf("[&(*it):0x%08X][it->a:%d][&(*vec.rend()):0x%08X]\n",
&(*it), it->a, &(*vec.rend()));
}


result:

[&(*it):0x01ADC010][it->a:0][&(*vec.rend()):0x01ADC028]
[&(*it):0x01ADC008][it->a:33][&(*vec.rend()):0x01ADC028]
[&(*it):0x01ADC000][it->a:0][&(*vec.rend()):0x01ADC048]


If vector has 5 elements, it works normally.

[&(*it):0x007620A0][it->a:4][&(*vec.rend()):0x00762078]
[&(*it):0x00762098][it->a:3][&(*vec.rend()):0x00762078]
[&(*it):0x00762090][it->a:2][&(*vec.rend()):0x00762078]
[&(*it):0x00762088][it->a:1][&(*vec.rend()):0x00762078]
[&(*it):0x00762080][it->a:0][&(*vec.rend()):0x00762078]
---------------------------------------------
[&(*it):0x007620A8][it->a:13][&(*vec.rend()):0x00762078]
[&(*it):0x007620A0][it->a:4][&(*vec.rend()):0x00762078]
[&(*it):0x00762098][it->a:33][&(*vec.rend()):0x00762078]
[&(*it):0x00762090][it->a:2][&(*vec.rend()):0x00762078]
[&(*it):0x00762088][it->a:1][&(*vec.rend()):0x00762078]
[&(*it):0x00762080][it->a:0][&(*vec.rend()):0x00762078]

Answer

push_back invalidates iterators when it causes size to exceed capacity:

If the new size() is greater than capacity() then all iterators and references (including the past-the-end iterator) are invalidated. Otherwise only the past-the-end iterator is invalidated.

Basically, if you must push_back, make sure to reserve ahead of time so you don't invalidate your iterator.

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