Is this all numbers?

How do i find out if argv[1] is all numbers?

[parksie]

Loop through it, seeing if the characters are in the range '0' to '9'. You may need to include '-' for the first character.

Could you supply some code?

Ok this is what i came up with but it doesnt seem to work...

int Digit(char * test)
{
	for(int i = 0; i > strlen(test); i++)
	{
		if(isdigit(test[i]) == 0)
		{
			return 1;
		}
	}

	return 0;
}

[HarryW]

Think you lost your [ i ] there.

I think it was meant to be like this:

int Digit(char * test)
{
	for(int i = 0; i > strlen(test); i++)
	{
		if(isdigit(test[ i ]) == 0)
		{
			return 1;
		}
	}

	return 0;
}

I had the i their, but it made it italics instead.

[parksie]

Try this:

bool Digit(char *test) {
    for(int j = 0; j < strlen(test); j++) {
        if(isdigit(test[j]) == 0) {
            return true;
        }
    }

    return false;
}

I think you got your comparison mixed up between > and <.

[Sam Finch]

I thought I'd try out my asm skills on this one, try this

bool StringNumeric(char* MyString)
{
	__asm
	{
		push ebx
		push ecx
		push ax


		mov ax, 0x2dfe	;ah = '-' al = ('.' - 0x30)
		mov ebx, MyString
		mov ecx, 0		;ch must be zero for the jcxz instruction to work (ecx is 0 as we must add it to ebx)

		sub ah, [ebx]	;subtract the first character from ah (so ah = 0 if the string begins with a '-')
		sete cl			;if the 1st character is '-' ecx = 1, else ecx = 0
		add ebx, ecx	;adds ecx to ebx
		

loopstrt:

		mov cl, [ebx]	;store the curren character in cl
		inc ebx			;move to the next character
		jcxz stringfinished		;if we find a null char exit the loop

		sub cl, 0x30	;cl should be between 30 and 39, so subtract 30 and check it's below 9
		cmp cl, 9
		jb loopstrt

		sub al, cl		;compare cl to al (true if cl was '.' originally, set al to 0 if true
		jz loopstrt	;(if it's false al could be anything but we return false anyway

stringfinished:
		pop ax

		cmp cl, 0	;if cl = 0 return true else return false
		sete al		

		pop cx
		pop ebx
	}
}

i reckon that's probably the fastest way of doing it.