view src/std.c @ 21:7c2adb65ceac

Started working on dynamic memory I have a slightly adapted version of dynamic memory allocation from the H&R book in the program now, I will soon be adding the free function. I fixed sPrint so it will work with large buffers.
author Jonathan Pevarnek <pevarnj@gmail.com>
date Tue, 15 Mar 2011 22:55:36 -0400
parents 525eab23e68a
children 3fb0ec050ff3
line wrap: on
line source

#include <std.h>

char* itoa(int n, char *a)
{
	char *ret = a;
	if(n < 0) {
		*a++ = '-';
		n *= -1;
	}
	char *b = a;
	if(!n) *b++ = '0';
	for(; n; b++) {
		*b = n%10 + '0';
		n = n/10;
	}
	*b-- = '\0';
	for(; a < b; a++, b--) { //reverse
		char temp = *b;
		*b = *a;
		*a = temp;
	}
	return ret;
}

char* ftoa(double x, char *a, unsigned int prec)
{
	char *ret = a;
	int n = (int) x; //integer part
	double d = x - (double) n; //fractional part;
	itoa(n, a);
	if(prec) { //only do the decimal part if decimal parts were asked for
		while(*a && *++a); //get to the null character from itoa
		int i; //counter variable for the for loop
		*a++ = '.'; //put the decimal in place
		for(i = 0; i < prec; i++) {
			d = d*10; //the integer part is the decimal digit
			*a++ = ((int) d) + '0'; //add the integer part of d to the string
			d -= (int) d; //chop off the integer part
		} a--; //move back to the last digit
		while(*a != '.') {
			if(*a == '0') {
				a--;
				continue;
			} else if(*a > '0' && *a <= '9') {
				a++;
				break;
			}
		}
		*a = '\0';
	}
	return ret;
}

int atoi(char *a)
{
	short neg = 0;
	int n = 0;
	if(*a == '-') {
		neg = 1;
		a++;
	} else if(*a == '+') a++;
	while(*a >= '0' && *a <= '9')
		n = n*10 + (*a++ - '0');
	if(neg) n *= -1;
	return n;
}

double atof(char *a)
{
	int n = atoi(a);
	double x = 0;
	double dec = .1;
	while(*a != '.') {
		if(!(*a >= '0' && *a <= '9')) return n;
		a++;
	} a++; //a will be immediately after the decimal point
	while(*a >= '0' && *a <= '9') { //goes through the decimal part
		x += (*a - '0')*dec;
		dec *= .1;
		a++;
	}
	return n + x;
}

void sPrint(char *a)
{
	char *b = a;
	while(*b && *++b);
	do {
		putline(a, (b - a > CON_LEN)?CON_LEN:(b - a));
		a += CON_LEN;
	} while(a < b);
}

int strcmp(const char *a, const char *b)
{
	while(1) {
		if(*a - *b) return *a - *b;
		if(*a == '\0') return 0;
		a++;
		b++;
	}
	return -1;
}

char* sGet(char *a, unsigned int n)
{
	int length = getline(a, n);
	a[(length < n)?length:n - 1] = '\0';
	return a;
}

int arrayLookup(char *text, const char array[][10], int last)
{
	int i;
	for(i = 0; i < last; i++)
		if(!strcmp(array[i], text)) return i;
	return last;
}

char* append(char *dest, char *src)
{
	char *ret = dest;
	while(*dest&& *++dest); //get to null in first string
	while((*dest++ = *src++));
	return ret;
}

static Block *allocp = NULL; //the location of the last item allocated by malloc

void* malloc(size_t size)
{
	unsigned int nBlocks = 1 + (size + sizeof(Block) - 1)/sizeof(Block);
	//We are allocating a number of blocks equal to the number needed rounded up
	//plus one for the actual info block
	Block *cur, *prev;
	if(allocp == NULL) {
		allocp = (void*)0x200000; //this is where memory should start
		allocp->s.next = allocp;
		allocp->s.size = (0x1400000/sizeof(Block)); //I am going to limit the heap to 20MB for now
	}
	prev = allocp;
	for(cur = prev->s.next;; ) {
		if(cur->s.size >= nBlocks) { //this section is pretty much exactly copied from K&R
			if(cur->s.size == nBlocks) {
				prev->s.next = cur->s.next;
			} else {
				cur->s.size -= nBlocks;
				cur += cur->s.size;
				cur->s.size = nBlocks;
			}
			allocp = prev;
			return (void*)(cur + 1);
		} else if(cur == allocp) { //We went back to the start...
			return NULL;
		}
	}
}

/*
void free(void *ptr)
{
}
*/