confused: vector<char*> and malloc

[Richard]

vector<char*> m_Text;
m_Text.resize(1 );
char* foo = "FOO";
char* bar = "BAR";
char* foobar = (char*)malloc(s trlen(foo) + strlen(bar) + 1);
if (foobar)
{
strcpy(foobar, foo);
strcat(foobar, bar);
}
m_Text[0] = foobar;

// Will m_Text[0] get freed when m_Text goes out of scope? If not, should I
call free(m_Text[0]) in the destructor?

[Victor Bazarov]

Richard wrote:

vector<char*> m_Text;
m_Text.resize(1 );
This makes 'm_Text' to be 1 element long. And since 'm_Text' was empty
prior to that, it adds 1 pointer to it and makes it null.
char* foo = "FOO";
char* bar = "BAR";
char* foobar = (char*)malloc(s trlen(foo) + strlen(bar) + 1);
if (foobar)
{
strcpy(foobar, foo);
strcat(foobar, bar);
}
At this point 'foobar' is a pointer to 7 character array, allocated in
the free store. The array has 'F', 'O', 'O', 'O', 'B', 'A', 'R', '\0'
in it, in sequence.
m_Text[0] = foobar;
This makes the value of the only element in the vector 'm_Text' to be the
same as the pointer to that 7-character array.
// Will m_Text[0] get freed when m_Text goes out of scope?
No.
If not, should I
call free(m_Text[0]) in the destructor?

Probably. Assuming you're talking about the destructor of the class in
which 'm_Text' is a data member.

V

[Kyle]

Richard wrote:

vector<char*> m_Text;
m_Text.resize(1 );
char* foo = "FOO";
char* bar = "BAR";
char* foobar = (char*)malloc(s trlen(foo) + strlen(bar) + 1);
malloc is C, why dont you use new ?
if (foobar)
{
strcpy(foobar, foo);
strcat(foobar, bar);
}
m_Text[0] = foobar;

// Will m_Text[0] get freed when m_Text goes out of scope? If not, should I
call free(m_Text[0]) in the destructor?

try this if you dont want to manage memory on your own

vector<string> m_Text;

char* foo = "FOO";
char* bar = "BAR";

string foobar(foo);
foobar.append( bar );

m_Text.push_bac k( foobar );

[Ali Çehreli]

"Kyle" <in*****@e.mail .com> wrote in message
news:dd******** **@nemesis.news .tpi.pl...

Richard wrote:

vector<char*> m_Text;
m_Text.resize(1 );
char* foo = "FOO";
char* bar = "BAR";
char* foobar = (char*)malloc(s trlen(foo) + strlen(bar) + 1);
malloc is C, why dont you use new ?

malloc is C++ too.

It is used to allocate raw memory in C++. Since new (and new[]) allocates
space and constructs object(s), new (and new[]) is not suitable when there
is not enough information to construct the object(s).
try this if you dont want to manage memory on your own

vector<string> m_Text;

Great advice!

Ali

[Richard]

"Kyle" <in*****@e.mail .com> wrote in message
news:dd******** **@nemesis.news .tpi.pl...

Richard wrote:

vector<char*> m_Text;
m_Text.resize(1 );
char* foo = "FOO";
char* bar = "BAR";
char* foobar = (char*)malloc(s trlen(foo) + strlen(bar) + 1);

malloc is C, why dont you use new ?

if (foobar)
{
strcpy(foobar, foo);
strcat(foobar, bar);
}
m_Text[0] = foobar;

// Will m_Text[0] get freed when m_Text goes out of scope? If not, should I call free(m_Text[0]) in the destructor?

try this if you dont want to manage memory on your own

vector<string> m_Text;

char* foo = "FOO";
char* bar = "BAR";

string foobar(foo);
foobar.append( bar );

m_Text.push_bac k( foobar );

That leads me to another question. Is it common practice to pass std::string
as a function argument? Is there much overhead?

void Foo(string& str)

If I went with vector<string>, would I need to covert all my functions that
currently use char* as an argument?

Thanks.

[Victor Bazarov]

Richard wrote:

[...] Is it common practice to pass std::string
as a function argument?
By value, no. By a reference, or by a reference to const, yes.
Is there much overhead?
There can be. Just like with any other UDT.
void Foo(string& str)

If I went with vector<string>, would I need to covert all my functions that
currently use char* as an argument?

Yes, most likely. BTW, if your functions that currently use 'char*' do
not change the characters in the arrays, you should declare 'char const*'
as the argument type. Then you don't have to change much, but you will
need to use 'c_str' member:

void my_function(cha r const*);
...
my_function(v[i].c_str());

V

[Richard]

"Victor Bazarov" <v.********@com Acast.net> wrote in message
news:X%******** ***********@new sread1.mlpsca01 .us.to.verio.ne t...

Richard wrote:

[...] Is it common practice to pass std::string
as a function argument?

By value, no. By a reference, or by a reference to const, yes.

> Is there much overhead?

There can be. Just like with any other UDT.

void Foo(string& str)

If I went with vector<string>, would I need to covert all my functions that currently use char* as an argument?

Yes, most likely. BTW, if your functions that currently use 'char*' do
not change the characters in the arrays, you should declare 'char const*'
as the argument type. Then you don't have to change much, but you will
need to use 'c_str' member:

void my_function(cha r const*);
...
my_function(v[i].c_str());

V

My array of strings is very large. I did a small test using char* vs string
and the results were really bad. I must be doing something wrong:

vector<char*> test;
test.resize(100 00000);
for(int t = 0; t != 10000000; ++t)
{
test[t] = "TEST 123";
}

That takes up a minimal amount of memory if using char*. But if you change
it to vector<string> test; it takes up around 1000 times more memory! What
am I doing wrong?

[Victor Bazarov]

Richard wrote:

My array of strings is very large. I did a small test using char* vs string
and the results were really bad. I must be doing something wrong:

vector<char*> test;
test.resize(100 00000);
for(int t = 0; t != 10000000; ++t)
{
test[t] = "TEST 123";
}

That takes up a minimal amount of memory if using char*.
Yeah... You got a vector all elements of which are the same. No extra
memory is allocated, just 10 million pointers. The consumption of memory
(not considering the overhead for dynamic memory management) is quite easy
to calculate:

sizeof(vector<c har*>) +
sizeof(char*) * test.size() + sizeof("TEST 123")

(which should give about 40000000, give or take a few bytes)
But if you change
it to vector<string> test; it takes up around 1000 times more memory! What
am I doing wrong?

I am not sure, to be honest with you. Every 'string' maintains its own
array of char internally. When you resize the 'test' vector to contain
ten millions of 'string' objects, it first puts a bunch of empty ones
there, and then when you in the loop assign those values, every string
in the vector needs to allocate its own small array (and possibly a bit
larger than asked for), which may lead to severe fragmentation of memory,
especially considering that a temporary is probably created to accommodate
your "TEST 123" literal... The final memory cost should be around

sizeof(vector<s tring>) +
sizeof(string) * test.size() +
sizeof("TEST 123") * test.size()

It is hard to believe it takes "around 1000 times more memory". The
string objects themselves are not that big, so you might be looking at
four, maybe ten, times the memory consumption, but definitely not the
thousand times. Are you running on a 64-bit machine? 1000 times more
with a vector of 10 million pointers is beyond what a 32-bit machine can
handle, that's for sure.

V

[Default User]

Richard wrote:

My array of strings is very large. I did a small test using char* vs
string and the results were really bad. I must be doing something
wrong:

vector<char*> test;
test.resize(100 00000);
for(int t = 0; t != 10000000; ++t)
{
test[t] = "TEST 123";
}

That takes up a minimal amount of memory if using char*. But if you
change it to vector<string> test; it takes up around 1000 times more
memory! What am I doing wrong?

You stuff 100 million copies of a pointer to the same piece of memory
into the vector. In the real application, you would presumably have a
different string in each slot in the vector.

To do a fairer comparison:

vector<char*> test;
test.resize(100 00000);
char* tmp = "TEST 123";

for(int t = 0; t != 10000000; ++t)
{
test[t] = new char[9];
strcpy(test[t], tmp);
}


Brian

[Nick Keighley]

Victor Bazarov wrote:

Richard wrote:

vector<char*> m_Text;
m_Text.resize(1 );
<snip>
char* foo = "FOO";
char* bar = "BAR";
char* foobar = (char*)malloc(s trlen(foo) + strlen(bar) + 1);
if (foobar)
{
strcpy(foobar, foo);
strcat(foobar, bar);
}

At this point 'foobar' is a pointer to 7 character array, allocated in
the free store. The array has 'F', 'O', 'O', 'O', 'B', 'A', 'R', '\0'
in it, in sequence.

'F', 'O', 'O', 'B', 'A', 'R', '\0'

<snip>