warning C4018: '<' : signed/unsigned mismatch ON VECTORS

Question

0.00/5 (No votes)

See more:

Hi,

when i run my program at the for loop i receive a warning which says:

warning C4018: '<' : signed/unsigned mismatch

C++

void insertion_sort(vector<int> &items)
{
	cout<<" insertion_sort... ";
	int key = 0;
	int i=0;
	for( int j=2; j<items.size(); j++){//HERE
		key = items[j];
		i = j-1;
		while (i>0 && items[i]>key){
			items[i+1] = items[i];
			i=i-1;
		}
		items[i+1] = key;
	}
}

it seems like i cannot use vector.size() .
comparing a signed or unsigned int with a vector.size(), vector.max_size() won't work. i can't even:

C++

cout<<items.size();
//or
cout<<items.max_size();
//or
signed int a = items.size();
cout<<a;
//or
unsigned int a = items.size();
cout<<a;
//or
int a = items.size();
cout<<a;

Why does it happen?
How can i fix it?
Thanks,

Posted 26-Jul-14 23:22pm

m.r.m.40

Updated 26-Jul-14 23:56pm

v3

Add a Solution

Comments

Mohibur Rashid 29-Jul-14 20:55pm

declare your j as size_t which is a typedef of unsigned int.

3 solutions

Add a Solution

Add your solution here

Treat my content as plain text, not as HTML

Preview 0

…

Existing Members

Sign in to your account

...or Join us

Download, Vote, Comment, Publish.

Your Email
Password
Forgot your password?

Your Email
This email is in use. Do you need your password?
Optional Password

I have read and agree to the Terms of Service and Privacy Policy
Please subscribe me to the CodeProject newsletters

When answering a question please:

Read the question carefully.
Understand that English isn't everyone's first language so be lenient of bad spelling and grammar.
If a question is poorly phrased then either ask for clarification, ignore it, or edit the question and fix the problem. Insults are not welcome.
Don't tell someone to read the manual. Chances are they have and don't get it. Provide an answer or move on to the next question.

Let's work to help developers, not make them feel stupid.

This content, along with any associated source code and files, is licensed under The Code Project Open License (CPOL)

Graham Breach · Answer 1 · 2014-07-26T23:34:00

Solution 1

That's because you declared j as an int, but items.size() returns an unsigned value. Declaring i and j as unsigned int should make it go away.

Posted 26-Jul-14 23:34pm

Graham Breach

Comments

m.r.m.40 27-Jul-14 5:36am

it didn't change anything:

unsigned int i=0;
for(unsigned int j=2; j

Stefan_Lang 29-Jul-14 6:36am

items.size() returns a std::size_t, which is usally defined as a 64 bit unsigned int, not a (usually 32 bit) unsigned int. While your suggestion should work normally, it doesn't address the case where items.size() > 2^32. Of course, that may be well out of practical range :-)

Graham Breach 29-Jul-14 7:29am

I'm sure you're right about it being a size_t in the end, but the header file has it returning a size_type member typedef. I was trying to keep my answer simple, my thinking being that "declare i and j as vector<int>::size_type instead of int" could have caused even more confusion!

Thomas Daniels · Answer 2 · 2014-07-26T23:35:00

Solution 2

items.size() returns a std::size_t, usually a 64-bit unsigned int, and j is a signed int. The compiler warns you for that, because you compare j with items.size(). Make j and i an unsigned long (an unsigned 64-bit int).

C++

unsigned long i=0;
for(unsigned long j=2; j<items.size(); j++){

Posted 26-Jul-14 23:35pm

Thomas Daniels

Updated 29-Jul-14 0:48am

v3

Comments

m.r.m.40 27-Jul-14 5:48am

it didn't work.
this time without any warning or error i get a message:
assertion failed.

is there any other way that i can get the size or index of the last item of the vector ?

Thomas Daniels 27-Jul-14 5:51am

It looks like the assertion happens in another part of your code, which we cannot see, so we cannot know the reason.

m.r.m.40 27-Jul-14 5:52am

I'll update the question.

Thomas Daniels 27-Jul-14 5:59am

Your updated question does not contain anything that looks like assertion; try to find where the assert function is used; this method causes the 'assertion failed'.

Stefan_Lang 29-Jul-14 6:31am

incorrect: items.size() returns a std::size_t, which is usally defined as a 64 bit unsigned int, not a (usually 32 bit) unsigned int. There are a few systems where those two types are the same, but they are rather rare, and I wouldn't assume that the author is using one.

Thomas Daniels 29-Jul-14 6:48am

Thanks for pointing that out! I edited my answer.

JJMatthews · Answer 3 · 2014-07-28T05:12:00

Solution 3

#pragma warning ( disable : 4018 )

or

void insertion_sort(vector<int> &items)
{
	cout<<" insertion_sort... ";
	int key = 0;
	int i=0;
	for( int j=2; j<(int)items.size(); j++){//HERE
		key = items[j];
		i = j-1;
		while (i>0 && items[i]>key){
			items[i+1] = items[i];
			i=i-1;
		}
		items[i+1] = key;
	}
}</int>

Posted 28-Jul-14 5:12am

JJMatthews

v2

Comments

Stefan_Lang 29-Jul-14 6:41am

Voted a 2. Reason: both suggestion do nothing to fix the problem and pretend that everything is ok. but it isn't. Even worse, the second suggestion involves a C-style type cast in a C++ program. Please don't do that! Type casts are signs of resignation. C-style type casts casts are signs of ignorance. I suggest you read up on casting here[^].

Note that both suggestions will likely fail if your vector contains more than 2^31 items (roughly two billion). And that is exactly what the warning is about! A std::vector actually can contain that many items - in fact it can contain many more, provided you compile the program as 64 bit. (Of course, with that many items, no one would live to tell whether or not the insertion sort worked all right - but that's another issue...)

m.r.m.40 30-Jul-14 5:41am

Well thank you very much for your great tips.

JJMatthews 29-Jul-14 11:31am

... just giving the learning individual 2 other possible solutions (which I use every day). I'm not saying it's the best solution but why not be aware of the available options. There is absolutely nothing wrong with type casting, throwing words around like "ignorance" is ridiculous. If a person had a requirement to hold 2 billion items in an array (which hardly ever happens) than they would code accordingly.

Stefan_Lang 30-Jul-14 8:36am

A warning is an indicator that the code you've written isn't entirely clear about some of the processing steps, or may contain an issue that may or may not cause run time problems.

While in this particular piece of code these issues will very likely never cause an actual problem, a similar piece of code may very well break: E. g. when you implement the Sieve Of Erastosthenes algorithm, you may want to use a very large bit vector, to hold flags for the prime numbers you've found. That vector may very well contain more elements than a 32 bit integer tye can count. Neither of your suggestions will prevent that program from breaking at runtime, and it will take some time just to find the cause.

With regards to type casts, if you must use them, you should use the C++ cast operators, not C-style cast.

However, the vast majority of type casts in C++ programs shouldn't be there, and wouldn't be necessary if the program would be using the correct types for its variables!

The code here is a good example: the variables i and J should be able to index any element in the vector, so they must be able to hold any non-negative number up to the maximum number of elements in a vector. int or unsigned int may or may not be sufficient, but the only way to be sure is use the types specifically provided for this, i. e. std::size_t, or std::vector::sizetype.

You can of course choose to simply suppress the warning by either of the methods you suggested, but doing so is the same as "killing the messenger": you did nothing to fix the actual problem.

JJMatthews 30-Jul-14 15:12pm

In my 15 years writing C++ code professionally I thankfully have never had the pleasure of working with this level of nitpicking. I hold myself to much higher coding standards than has ever been required by any of my employers. Please don't discourage this young man into thinking this will be the way it is for him in his career because it is not.

let me make this simple for you:

warning != error;
warning != "problem";
warning == warning;