C / C++ / MFC

Re: How to find the positions of '\n' within a string object

21-Oct-10 9:30

What is "string object"? CString? std::string?

If it is CString, then as you can see in the documentation of CString::Find[^], you can specify it the starting character it will search from, so when n contains your first '\n', just feed find with n+1...

> The problem with computers is that they do what you tell them to do and not what you want them to do. <
> Leela: Fry, you're wasting your life sitting in front of that TV. You need to get out and see the real world.
Fry: But this is HDTV. It's got better resolution than the real world <

Alain Rist21-Oct-10 11:48

Passing a char array to a routine

21-Oct-10 11:48

TCPMem wrote:
but what if I want the second '\n'?

Use the (incremented) last position as start for the next find():

std::string str = "Line1Line1\nLine2Line2\n";
for (size_t pos = 0; pos != str.npos; pos = str.find('\n', ++pos))	
	std::cout << str.substr(pos); // do sompething with pos

cheers,
AR

When the wise (person) points at the moon the fool looks at the finger (Chinese proverb)

Andy20221-Oct-10 8:43

Andy202

21-Oct-10 8:43

I need to pass an array into a procedure and return a string. The problem is the array could be 1, 2, 3 or 4 character strings.

I am stuck as how to do it, see typical arrays below.

char *OPTIONS_ENUMS[4] = {
			"OPT_1",
			"OPT_2",
			"OPT_3",
			"OPT_4"};

			
char *OPTIONS2_ENUMS[4] = {
			"OPT_1",
			"OPT_4"};
		
			
char* GetOptions(short nStatus, char* szOptions)
{
	int i, nOptions;
	// How can I tell how many elements in the array?
	nOptions = sizeof(ssmStatusOptions);	// This will not work
	for(i=0; i < nOptions; i++)
	{
		if(nStatus == 0)
		{
			if(strcmp(szOptions[i], "NULL") == 0)
			{
				return "NULL";
			}
			else if(strcmp(szOptions[i], "ITEM1") == 0)
			{
				return "ITEM1";
			}
			else if(strcmp(szOptions[i], "NO_DATA") == 0)
			{
				return "NO_DATA";
			}
		}
		else if(nStatus == 1)
		{
			if(strcmp(szOptions[i], "OK") == 0)
			{
				return "OK";
			}
			else if(strcmp(szOptions[i], "ITEM2") == 0)
			{
				return "ITEM2";
			}
			else if(strcmp(szOptions[i], "NO_DATA") == 0)
			{
				return "NO_DATA";
			}
		}
	}
}

How to call the procedure

e.g.

GetOptions(1, OPTIONS2_ENUMS);

where OPTIONS2_ENUMS is

char *OPTIONS_ENUMS[4] = {
        "OPT_1",
        "OPT_2",
        "OPT_3",
        "OPT_4"};

I need to know how many elements in the array and how to pass over the array. Its a c code project.

Many thanks.

Code-o-mat21-Oct-10 9:27

21-Oct-10 9:27

You could either specify the array size to the function as an additional parameter, or you could terminate your array with NULL and in the function iterate until you get a NULL.
Btw, shouldn't this:

Andy202 wrote:
char* GetOptions(short nStatus, char* szOptions)

be instead this:

char* GetOptions(short nStatus, char **szOptions)

Andy20221-Oct-10 10:53

Andy202

21-Oct-10 10:53

Thanks for your suggestion. But are they anyway I can find out the number of elements in an array like this:-

char *OPTIONS_ENUMS[4] = {"OPT_1","OPT_2","OPT_3","OPT_4"};

// Would this work

for(int j = 0; j < 100; j++)
{
   if(OPTIONS_ENUMS[j] == NULL)
     return j;
}

Richard MacCutchan21-Oct-10 11:16

Richard MacCutchan

21-Oct-10 11:16

No, this will (probably) fail; you only have 4 elements in your array so whatever exists at the next word OPTIONS_ENUMS[4] is not guaranteed, it may or may not be NULL. It is a simple matter to declare an array like this:

char *OPTIONS_ENUMS[] = {"OPT_1","OPT_2","OPT_3","OPT_4", NULL};

and let the compiler work out how big it needs to be.

Just say 'NO' to evaluated arguments for diadic functions! Ash

Garth J Lancaster21-Oct-10 11:25

Garth J Lancaster

21-Oct-10 11:25

I agree with Richard's answer - but in some cases, it may not make sense to use NULL as a sentinal marking end of the list, sometimes you may end up defining a non NULL sentinal ..

for example, in your list, as long as "EOL" or "End-List" we guaranteed never to be part of your data, you could look through the list until "EOL" or "End-List" were found

char *OPTIONS_ENUMS[4] = {"OPT_1","OPT_2","OPT_3","OPT_4", "EOL"};

but I'd stick with NULL unless you had a good reason (eg, NULL could be part of the data if you were specifying an empty option/slot)

'g'

Re: Passing a char array to a routine [modified]

Richard MacCutchan21-Oct-10 11:59

Richard MacCutchan

21-Oct-10 11:59

See my comments to Alain.

Just say 'NO' to evaluated arguments for diadic functions! Ash

Alain Rist21-Oct-10 11:34

21-Oct-10 11:34

No it will crash, but Smile | :)

char *OPTIONS_ENUMS[5] = {"OPT_1","OPT_2","OPT_3","OPT_4"}; // This works
for(int j = 0; j < 100; j++)
{   
    if(OPTIONS_ENUMS[j] == NULL)     
    return j;
}

EDIT:
As Richard says in next post this may be error prone and is more a joke than a solution.

With C++0x (gcc 4.5 only for now) we can use the standard collections with initializer lists like:

std::vector<LPCSTR> OPTIONS_ENUMS = {"OPT_1","OPT_2","OPT_3","OPT_4"};

When not available (MS compilers for now) we must stay in C world and use a macro:

#define ARRAY_SIZE(_Array) (sizeof(_Array) / sizeof(_Array[0])) // This is the answer 

char *OPTIONS_ENUMS[] = {"OPT_1","OPT_2","OPT_3","OPT_4"}; 
for (j = 0; j < ARRAY_SIZE(OPTIONS_ENUMS); j++)
{   
    if(OPTIONS_ENUMS[j] == NULL) // will never happen   
    return j;

    // do something with OPTIONS_ENUMS[j]
}

cheers,
AR

When the wise (person) points at the moon the fool looks at the finger (Chinese proverb)

modified on Thursday, October 21, 2010 6:50 PM

Richard MacCutchan21-Oct-10 11:59

Richard MacCutchan

21-Oct-10 11:59

Both yours and Garth's answers have the array size specified in the declaration, which I always feel is prone to errors in the future. If you add or remove an element or elements and forget to amend the size then your program starts misbehaving. Stick with my solution and let the compiler do the work for you.

Just say 'NO' to evaluated arguments for diadic functions! Ash

Alain Rist21-Oct-10 12:37

21-Oct-10 12:37

It was more a joke than an answer, I edited it to a real answer.
cheers,
AR

When the wise (person) points at the moon the fool looks at the finger (Chinese proverb)

Andy20222-Oct-10 2:03

Andy202

22-Oct-10 2:03

Thanks I just tested this and it did not work. The size is always 1.

Have I implemented it correct?

#define ARRAY_SIZE(_Array) (sizeof(_Array) / sizeof(_Array[0])) // This is the answer

char *OPTIONS1_ENUMS[4] = {"OPT_1", "OPT_2", "OPT_3", "OPT_4"};
char *OPTIONS2_ENUMS[2] = {"OPT_1", "OPT_4"};
char *OPTIONS3_ENUMS[1] = {"OPT_1234"};
 
char* GetOptions(short enumValue, char** szOptions)
{
	int i, nOptions;
	short size;
	nOptions = 1;
	size = ARRAY_SIZE(szOptions);
	
	printf("Array Size = %d\n", size);
    // do something with OPTIONS_ENUMS[j]
}

// In the Main program call GetOptions() for 3 different sized arrays
GetOptions(1, OPTIONS1_ENUMS);
GetOptions(2, OPTIONS2_ENUMS);
GetOptions(3, OPTIONS3_ENUMS);

Alain Rist22-Oct-10 4:05

22-Oct-10 4:05

Andy202 wrote:
Thanks I just tested this and it did not work. The size is always 1.

Have I implemented it correct?

Nope

Use it as follows:

#define ARRAY_SIZE(_Array) (sizeof(_Array) / sizeof(_Array[0])) // This is the answer

char *OPTIONS1_ENUMS[] = {"OPT_1", "OPT_2", "OPT_3", "OPT_4"};
char *OPTIONS2_ENUMS[] = {"OPT_1", "OPT_4"};
char *OPTIONS3_ENUMS[] = {"OPT_1234"}; 

void GetOptions(short enumValue, char** szOptions, int Size)
{	
	printf("Array Size = %d\n", Size);    
}
int main()
{
	// In the Main program call GetOptions() for 3 different sized arrays
	GetOptions(1, OPTIONS1_ENUMS, ARRAY_SIZE(OPTIONS1_ENUMS));
	GetOptions(2, OPTIONS2_ENUMS, ARRAY_SIZE(OPTIONS2_ENUMS));
	GetOptions(3, OPTIONS3_ENUMS, ARRAY_SIZE(OPTIONS3_ENUMS));
}

BTW you should not call your arrays OPTIONSx_ENUMS:
- as capitalized names are conventionnally reserved for C macros, and
- they are not enums but C arrays of const char*.
cheers,
AR

When the wise (person) points at the moon the fool looks at the finger (Chinese proverb)

Code-o-mat21-Oct-10 21:40

21-Oct-10 21:40

Sorry for the late answer (went to bed, was night here), i agree with the others, use NULL or some special string (or a special pointer) to specify the end of the array.

Btw, you can also do this:

char *OPTIONS_ENUMS[] = {"OPT_1","OPT_2","OPT_3","OPT_4", NULL};

this way you don't have to specify the array size, your array's size will be determined by the items you enumerated.

is there something faster than GetPixel and SetPixel ? [moved]

Aescleal21-Oct-10 21:31

Aescleal

21-Oct-10 21:31

I might be really lazy but I'd be tempted to do one of two things:

- Pass another parameter into the function that says how big the array is
- Bundle the pointer into a structure with the number of entries and use that

That way you don't have to worry as much about pointers to pointers and other things that can make you go cross eyed. You're also half way to adding an abstract data type for your options handling as well.

From reading your code I'm not sure you're sure you know what you want to achive. The way you've written is it's "return the first array member that matches one of these strings." You've also not got a return value for when you can't find anything. It's been a while since I've programmed in C but I'd have thought your compiler would have a whinge about that!

Cheers,

Ash

inayathussaintoori21-Oct-10 8:38

inayathussaintoori

21-Oct-10 8:38

hello
i am working on a project "Face Detection" in live video without OpenCv, EmguCV etc. so the first thing that I want to access the RGB values of each pixel and then set them back after some plus minus calculation. but the problem is that i can't use GetPixel and SetPixel methods because they are too slow. and i need heigh speed. in one second i have to process many frames. my question is that is there any other way to get the rgb values and set them back more faster than GetPixel and SetPixel, which full fill my requirements?
thaks

Re: is there something faster than GetPixel and SetPixel ? [moved]

Code-o-mat21-Oct-10 9:19