Vector element erase cause SIGSEGV

[Bin Chen]

Hi,

I don't know what's wrong with my program, it causes SIGSEGV:

#include <vector>

using namespace std;

int main(void)
{
vector<intv;
vector<int>::iterator iter;

int i,j,k;

i = 9;j = 10;k = 11;

v.push_back(i);
v.push_back(j);
v.push_back(k);

for (iter = v.begin();iter != v.end(); ++iter) {
printf("*********%d\n", *iter);
iter = v.erase(iter);
}

return 0;
}

Thanks.
ABAI

[Thomas Tutone]

Bin Chen wrote:

I don't know what's wrong with my program, it causes SIGSEGV:

#include <vector>

using namespace std;

int main(void)
{
vector<intv;
vector<int>::iterator iter;

int i,j,k;

i = 9;j = 10;k = 11;

v.push_back(i);
v.push_back(j);
v.push_back(k);

for (iter = v.begin();iter != v.end(); ++iter) {

Change the above line to:

for (iter = v.begin();iter != v.end();) {

printf("*********%d\n", *iter);
iter = v.erase(iter);
}

return 0;
}

std::vector::erase returns an iterator to the element immediately
following the one that was erased. So in effect, the expression "iter
= v.erase(iter)" causes iter to to be incremented. But your for loop
also increments iter. In other words, iter is incremented _twice_ each
time through the loop. The first time through the loop, iter begins
pointing to v[0] and ends pointing to v[2]. The second time through
the loop, iter begins pointing to v[2]. After printing the value of
v[2], the program erased v[2], and increments iter so that it is equal
to v.end(). Then, the program attempts to increment iter again (as
happens every time through your for loop), leading to undefined
behavior - in this instance, a seg fault.

To fix your program, change the line indicated above.

Best regards,

Tom

[Guang Han]

"Bin Chen дµÀ£º
"

Hi,

I don't know what's wrong with my program, it causes SIGSEGV:

#include <vector>

using namespace std;

int main(void)
{
vector<intv;
vector<int>::iterator iter;

int i,j,k;

i = 9;j = 10;k = 11;

v.push_back(i);
v.push_back(j);
v.push_back(k);

for (iter = v.begin();iter != v.end(); ++iter) {
printf("*********%d\n", *iter);
iter = v.erase(iter);

erase will return the next element after the iter. assuming you are
erasing the last element, then iter will be v.end (), and then in your
for loop you'll still ++iter, so it should be some errors.....

}

return 0;
}

Thanks.
ABAI

[Haro Panosyan]

If the question can be reformulated as how to fix,
then this is something you need to do

int Size = v.size();
for (int I = 0; I < Size; I++) {
printf("*********%d\n", *v.begin());
v.erase(v.begin());
}
If you want really to know why it is not working,
this may give you a hint:
for (iter = v.begin();iter != v.end(); ++iter) {
printf("BEF *********%d\n", *iter);
iter = v.erase(iter);
printf("AFT *********%d\n", *iter);
}

Erasing any element except the last causes the vector to
be "rearanged", and return value is iterator to next element.
Erasing the last element seems not to return v.end() in my tests,
but returns the last element and this could be because the pointer
to the last location is in memory.

-haro
-----------------------------------------
Sorry if this is not complete answer, I just thought it
might help you quickly in case you are not trying to deeply
understand what is going on.
Bin Chen wrote:

Hi,

I don't know what's wrong with my program, it causes SIGSEGV:

#include <vector>

using namespace std;

int main(void)
{
vector<intv;
vector<int>::iterator iter;

int i,j,k;

i = 9;j = 10;k = 11;

v.push_back(i);
v.push_back(j);
v.push_back(k);

for (iter = v.begin();iter != v.end(); ++iter) {
printf("*********%d\n", *iter);
iter = v.erase(iter);
}

return 0;
}

Thanks.
ABAI

[Nate Barney]

Thomas Tutone wrote:

>
for (iter = v.begin();iter != v.end();) {

printf("*********%d\n", *iter);
iter = v.erase(iter);
}

I would be tempted to write the loop this way:

for (iter = v.begin(); iter != v.end(); iter = v.erase(iter))
printf("*********%d\n", *iter);

Does anyone care to comment on whether or not this is good style?

--
Nate