State Machine of different Variables in C - c

I have the following code which is doing lot of if-else check and consuming lot of time.
while(i<1000) {
if ( a == x1)
{
if(b == y1)
{
array[i] = z1;
}
else if( b = y2)
{
array[i] = z2;
}
else if (b = z3);
{
array[i] = z3;
}
-
-
-
-
-
some 20 else --
}/*end of first if*/
else if (a == xx1)
{
if(b == yy1)
{
array[i] = zz1;
}
else (b == yyy3)
{
array[i] = zz2;
}
else ( b == yy4)
{
array[i] = zz3;
}
-
-
-
-
-
some 20 else --
}
else if(/*more conditions*/)
{
/* same as above with too many else loops */
}
-
-
-
-
-
-
-some 100 else-ifs
-
-
-
-
-
i++;
}/*end of while*/
How can i minimize the time consumption.I though of a replacing if else with switch(switch within the switch) , but still i think i can do better with
some tables(2d-3d array) but the numbers x1,y1,z1 ,zz3,zz1,zz2 are not continous numbers and have no relation among them they exists between 1 to 1 billion with no
repetetion.Any thoughts for implementing this lookup table.
Rgds,
softy

(1) If the numbers are constants change it to case statement - your compiler should output a jump table if it can (given your constraints it can't - so go to step 2).
(2) If your data is non-contiguous a jump table won't work. You can make your data contiguous (or close enough to it) by using a hash table. Now just search the hash table. This should be O(1) (or close to it). Though you'll have to bear the cost of creating the hash table
If you need code for a hash table you can use C++ STL or if you need C write your own or find an implementation on the web, this question might help.

Related

How would I go about adding definitions to my language?

I have written a stack based interpreted language in C. The interpreter works with reading a file line by line and executing the line depending on the operation. print(1 + 1) in python would become 1 1 + print.
Here is the function to check what the operation is and push it to the stack or do an operation on it:
if (strncmp(op, "+", 1) == 0 && is_definition == 0)
{
float a = pop(stack);
float b = pop(stack);
push(stack, a + b);
}
else if (strncmp(op, "-", 1) == 0 && is_definition == 0)
{
float a = pop(stack);
float b = pop(stack);
push(stack, b - a);
}
else if (strncmp(op, "*", 1) == 0 && is_definition == 0)
{
float a = pop(stack);
float b = pop(stack);
push(stack, a * b);
}
else if (strncmp(op, "/", 1) == 0 && is_definition == 0)
{
float a = pop(stack);
float b = pop(stack);
push(stack, b / a);
}
else if (strncmp(op, "print", 5) == 0 && is_definition == 0)
{
float a = pop(stack);
if(a == (int)a)
printf("%d\n", (int)a);
else
printf("%f\n", a);
}
else if (strncmp(op, "define", 6) == 0 && is_definition == 0)
{
is_definition = 1;
}
else if (is_definition == 1)
{
}
else if (strncmp(op, "end", 3) == 0 && is_definition == 1)
{
is_definition = 0;
}
else
{
push(stack, atoi(op));
}
This is inside a loop that iterates over every space separated operation in the code.
I want to add a definition system a bit like the one in C.
This is the syntax I would like to have
define TEST 10 end
I would like to use this a bit like a variable system where you can use TEST.
In pseudo code, you should do the following:
Read a line of source
If it is a definition, parse+store it and skip the rest
(it is not a definition) execute, much like the code you posted
About "parse+store" the definitions, you need - for example - a couple of arrays, or an array of structs. You need to store each "name" (the alias, or the name of the definition) and, along with each name, its value.
Then, in the code you posted, you should implement the push() instruction (you only mention pop()). The push() instruction reads an operand and determines if it is an alias (definition) or not:
(push pseudo code)
Get the operand
Determine if it is a definition. Basically, you iterate on all the stored definitions to find a correspondence
Got the final value, put it on the stack
There are several things that could be said... a couple of them, in sparse order:
The pushed operand is a number? In this case you can skip the definition(s) checking, assuming that it is illegal to say "define 10 20"
Would you allow to (re)define operators?
Would you allow a definition to refer to other definitions?
...

Can this weird function be directly calculated?

I'd like the calculate this function directly, but the trick has eluded me so far:
uint8_t distance(uint64_t a, uint8_t b) {
// a and b both odd, a at least as large as b
assert((a & 1) && (b & 1) && a >= b);
// really dumb, keep subtracting 2 until you hit a multiple of b
uint64_t distance = 0;
while (a % b != 0) {
distance++;
a -= 2;
}
assert(distance <= 255);
return (uint8_t)distance;
}
Basically the return value is how many times a has to be decremented by 2 to be a multiple of b. This should be true eventually (without wraparound or anything weird) since a and b are odd and a >= b.
If it was "decrement by 1" rather than by 2 the answer would be a simple %...
Performance matters in this case (so for example avoiding unpredictable branches would be nice).
If a % b is even, the result is (a % b) / 2. Otherwise, it's (a % b + b) / 2. You can microoptimize this if you want, but a straightforward implementation is
if ((a % b) % 2 == 0) {
return (a % b) / 2;
} else {
return (a % b + b) / 2;
}
I won't give you a specific microoptimized version, because that's something you should really profile yourself in the context of your application.

C : Sum of reverse numbers

So I want to solve an exercise in C or in SML but I just can't come up with an algorithm that does so. Firstly I will write the exercise and then the problems I'm having with it so you can help me a bit.
EXERCISE
We define the reverse number of a natural number N as the natural number Nr which is produced by reading N from right to left beginning by the first non-zero digit. For example if N = 4236 then Nr = 6324 and if N = 5400 then Nr = 45.
So given any natural number G (1≤G≤10^100000) write a program in C that tests if G can occur by the sum of a natural number N and its reverse Nr. If there is such a number then the program must return this N. If there isn't then the program must return 0. The input number G will be given through a txt file consisted only by 1 line.
For example, using C, if number1.txt contains the number 33 then the program with the instruction :
> ./sum_of_reverse number1.txt
could return for example 12, because 12+21 = 33 or 30 because 30 + 3 = 33. If number1.txt contains the number 42 then the program will return 0.
Now in ML if number1.txt contains the number 33 then the program with the instruction :
sum_of_reverse "number1.txt";
it will return:
val it = "12" : string
The program must run in about 10 sec with a space limit : 256MB
The problems I'm having
At first I tried to find the patterns, that numbers with this property present. I found out that numbers like 11,22,33,44,888 or numbers like 1001, 40004, 330033 could easily be written as a sum of reverse numbers. But then I found out that these numbers seem endless because of numbers for example 14443 = 7676 + 6767 or 115950 = 36987 + 78963.
Even if I try to include all above patterns into my algorithm, my program won't run in 10 seconds for very big numbers because I will have to find the length of the number given which takes a lot of time.
Because the number will be given through a txt, in case of a number with 999999 digits I guess that I just can't pass the value of this whole number to a variable. The same with the result. I assume that you are going to save it to a txt first and then print it??
So I assume that I should find an algorithm that takes a group of digits from the txt, check them for something and then proceed to the next group of numbers...?
Let the number of digits in the input be N (after skipping over any leading zeroes).
Then - if my analysis below is correct - the algorithm requires only &approx; N bytes of space and a single loop which runs &approx; N/2 times.
No special "big number" routines or recursive functions are required.
Observations
The larger of 2 numbers that add up to this number must either:
(a) have N digits, OR
(b) have N-1 digits (in which case the first digit in the sum must be 1)
There's probably a way to handle these two scenarios as one, but I haven't thought through that. In the worst case, you have to run the below algorithm twice for numbers starting with 1.
Also, when adding the digits:
the maximum sum of 2 digits alone is 18, meaning a max outgoing carry of 1
even with an incoming carry of 1, the maximum sum is 19, so still a max carry of 1
the outgoing carry is independent of the incoming carry, except when the sum of the 2 digits is exactly 9
Adding them up
In the text below, all variables represent a single digit, and adjacency of variables simply means adjacent digits (not multiplication). The ⊕ operator denotes the sum modulo 10. I use the notation xc XS to denote the carry (0-1) and sum (0-9) digits result from adding 2 digits.
Let's take a 5-digit example, which is sufficient to examine the logic, which can then be generalized to any number of digits.
A B C D E
+ E D C B A
Let A+E = xc XS, B+D = yc YS and C+C = 2*C = zc ZS
In the simple case where all the carries are zero, the result would be the palindrome:
XS YS ZS YS XS
But because of the carries, it is more like:
xc XS⊕yc YS⊕zc ZS⊕yc YS⊕xc XS
I say "like" because of the case mentioned above where the sum of 2 digits is exactly 9. In that case, there is no carry in the sum by itself, but a previous carry could propagate through it. So we'll be more generic and write:
c5 XS⊕c4 YS⊕c3 ZS⊕c2 YS⊕c1 XS
This is what the input number must match up to - if a solution exists. If not, we'll find something that doesn't match and exit.
(Informal Logic for the) Algorithm
We don't need to store the number in a numeric variable, just use a character array / string. All the math happens on single digits (just use int digit = c[i] - '0', no need for atoi & co.)
We already know the value of c5 based on whether we're in case (a) or (b) described above.
Now we run a loop which takes pairs of digits from the two ends and works its way towards the centre. Let's call the two digits being compared in the current iteration H and L.
So the loop will compare:
XS⊕c4 and XS
YS⊕c3 and YS⊕c1
etc.
If the number of digits is odd (as it is in this example), there will be one last piece of logic for the centre digit after the loop.
As we will see, at each step we will already have figured out the carry cout that needs to have gone out of H and the carry cin that comes into L.
(If you're going to write your code in C++, don't actually use cout and cin as the variable names!)
Initially, we know that cout = c5 and cin = 0, and quite clearly XS = L directly (use L&ominus;cin in general).
Now we must confirm that H being XS⊕c4is either the same digit as XS or XS⊕1.
If not, there is no solution - exit.
But if it is, so far so good, and we can calculate c4 = H&ominus;L. Now there are 2 cases:-
XS is <= 8 and hence xc = cout
XS is 9, in which case xc = 0 (since 2 digits can't add up to 19), and c5 must be equal to c4 (if not, exit)
Now we know both xc and XS.
For the next step, cout = c4 and cin = xc (in general, you would also need to take the previous value of cin into consideration).
Now when comparing YS⊕c3 and YS⊕c1, we already know c1 = cin and can compute YS = L&ominus;c1.
The rest of the logic then follows as before.
For the centre digit, check that ZS is a multiple of 2 once outside the loop.
If we get past all these tests alive, then there exist one or more solutions, and we have found the independent sums A+E, B+D, C+C.
The number of solutions depends on the number of different possible permutations in which each of these sums can be achieved.
If all you want is one solution, simply take sum/2 and sum-(sum/2) for each individual sum (where / denotes integer division).
Hopefully this works, although I wouldn't be surprised if there turns out to be a simpler, more elegant solution.
Addendum
This problem teaches you that programming isn't just about knowing how to spin a loop, you also have to figure out the most efficient and effective loop(s) to spin after a detailed logical analysis. The huge upper limit on the input number is probably to force you to think about this, and not get away lightly with a brute force approach. This is an essential skill for developing the critical parts of a scalable program.
I think you should deal with your numbers as C strings. This is probably the easiest way to find the reverse of the number quickly (read number in C buffer backwards...) Then, the fun part is writing a "Big Number" math routines for adding. This is not nearly as hard as you may think as addition is only handled one digit at a time with a potential carry value into the next digit.
Then, for a first pass, start at 0 and see if G is its reverse. Then 0+1 and G-1, then... keep looping until G/2 and G/2. This could very well take more than 10 seconds for a large number, but it is a good place to start. (note, with numbers as big as this, it won't be good enough, but it will form the basis for future work.)
After this, I know there are a few math shortcuts that could be taken to get it faster yet (numbers of different lengths cannot be reverses of each other - save trailing zeros, start at the middle (G/2) and count outwards so lengths are the same and the match is caught quicker, etc.)
Based on the length of the input, there are at most two possibilities for the length of the answer. Let's try both of them separately. For the sake of example, let's suppose the answer has 8 digits, ABCDEFGH. Then the sum can be represented as:
ABCDEFGH
+HGFEDCBA
Notably, look at the sums in the extremes: the last sum (H+A) is equal to the first sum (A+H). You can also look at the next two sums: G+B is equal to B+G. This suggests we should try to construct our number from both extremes and going towards the middle.
Let's pick the extremes simultaneously. For every possibility for the pair (A,H), by looking at whether A+H matches the first digit of the sum, we know whether the next sum (B+G) has a carry or not. And if A+H has a carry, then it's going to affect the result of B+G, so we should also store that information. Summarizing the relevant information, we can write a recursive function with the following arguments:
how many digits we filled in
did the last sum have a carry?
should the current sum have a carry?
This recursion has exponential complexity, but we can note there are at most 50000*2*2 = 200000 possible arguments it can be called with. Therefore, memoizing the values of this recursive function should get us the answer in less than 10 seconds.
Example:
Input is 11781, let's suppose answer has 4 digits.
ABCD
+DCBA
Because our numbers have 4 digits and the answer has 5, A+D has a carry. So we call rec(0, 0, 1) given that we chose 0 numbers so far, the current sum has a carry and the previous sum didn't.
We now try all possibilities for (A,D). Suppose we choose (A,D) = (9,2). 9+2 matches both the first and final 1 in the answer, so it's good. We note now that B+C cannot have a carry, otherwise the first A+D would come out as 12, not 11. So we call rec(2, 1, 0).
We now try all possibilities for (B,C). Suppose we choose (B,C) = (3,3). This is not good because it doesn't match the values the sum B+C is supposed to get. Suppose we choose (B,C) = (4,3). 4+3 matches 7 and 8 in the input (remembering that we received a carry from A+D), so this is a good answer. Return "9432" as our answer.
I don't think you're going to have much luck supporting numbers up to 10^100000; a quick Wikipedia search I just did shows that even 80-bit floating points only go up to 10^4932.
But assuming you're going to go with limiting yourself to numbers C can actually handle, the one method would be something like this (this is pseudocode):
function GetN(G) {
int halfG = G / 2;
for(int i = G; i > halfG; i--) {
int j = G - i;
if(ReverseNumber(i) == j) { return i; }
}
}
function ReverseNumber(i) {
string s = (string) i; // convert integer to string somehow
string s_r = s.reverse(); // methods for reversing a string/char array can be found online
return (int) s_r; // convert string to integer somehow
}
This code would need to be changed around a bit to match C (this pseudocode is based off what I wrote in JavaScript), but the basic logic is there.
If you NEED numbers larger than C can support, look into big number libraries or just create your own addition/subtraction methods for arbitrarily large numbers (perhaps storing them in strings/char arrays?).
A way to make the program faster would be this one...
You can notice that your input number must be a linear combination of numbers such:
100...001,
010...010,
...,
and the last one will be 0...0110...0 if #digits is even or 0...020...0 if #digits is odd.
Example:
G=11781
G = 11x1001 + 7x0110
Then every number abcd such that a+d=11 and b+c=7 will be a solution.
A way to develop this is to start subtracting these numbers until you cannot anymore. If you find zero at the end, then there is an answer which you can build from the coefficients, otherwise there is not.
I made this and it seems to work:
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
int Counter (FILE * fp);
void MergePrint (char * lhalf, char * rhalf);
void Down(FILE * fp1, FILE * fp2, char * lhalf, char * rhalf, int n);
int SmallNums (FILE * fp1, int n);
int ReverseNum (int n);
int main(int argc, char* argv[])
{
int dig;
char * lhalf = NULL, * rhalf = NULL;
unsigned int len_max = 128;
unsigned int current_size_k = 128;
unsigned int current_size_l = 128;
lhalf = (char *)malloc(len_max);
rhalf =(char *)malloc(len_max);
FILE * fp1, * fp2;
fp1 = fopen(argv[1],"r");
fp2 = fopen(argv[1],"r");
dig = Counter(fp1);
if ( dig < 3)
{
printf("%i\n",SmallNums(fp1,dig));
}
else
{
int a,b,prison = 0, ten = 0, i = 0,j = dig -1, k = 0, l = 0;
fseek(fp1,i,0);
fseek(fp2,j,0);
if ((a = fgetc(fp1)- '0') == 1)
{
if ((fgetc(fp1)- '0') == 0 && (fgetc(fp2) - '0') == 9)
{
lhalf[k] = '9';
rhalf[l] = '0';
i++; j--;
k++; l++;
}
i++;
prison = 0;
ten = 1;
}
while (i <= j)
{
fseek(fp1,i,0);
fseek(fp2,j,0);
a = fgetc(fp1) - '0';
b = fgetc(fp2) - '0';
if ( j - i == 1)
{
if ( (a == b) && (ten == 1) && (prison == 0) )
Down(fp1,fp2,lhalf,rhalf,0);
}
if (i == j)
{
if (ten == 1)
{
if (prison == 1)
{
int c;
c = a + 9;
if ( c%2 != 0)
Down(fp1,fp2,lhalf,rhalf,0);
lhalf[k] = c/2 + '0';
k++;
}
else
{
int c;
c = a + 10;
if ( c%2 != 0)
Down(fp1,fp2,lhalf,rhalf,0);
lhalf[k] = c/2 + '0';
k++;
}
}
else
{
if (prison == 1)
{
int c;
c = a - 1;
if ( c%2 != 0)
Down(fp1,fp2,lhalf,rhalf,0);
lhalf[k] = c/2 + '0';
k++;
}
else
{
if ( a%2 != 0)
Down(fp1,fp2,lhalf,rhalf,0);
lhalf[k] = a/2 + '0';
k++;
}
}
break;
}
if (ten == 1)
{
if (prison == 1)
{
if (a - b == 0)
{
lhalf[k] = '9';
rhalf[l] = b + '0';
k++; l++;
}
else if (a - b == -1)
{
lhalf[k] = '9';
rhalf[l] = b + '0';
ten = 0;
k++; l++;
}
else
{
Down(fp1,fp2,lhalf,rhalf,0);
}
}
else
{
if (a - b == 1)
{
lhalf[k] = '9';
rhalf[l] = (b + 1) + '0';
prison = 1;
k++; l++;
}
else if ( a - b == 0)
{
lhalf[k] = '9';
rhalf[l] = (b + 1) + '0';
ten = 0;
prison = 1;
k++; l++;
}
else
{
Down(fp1,fp2,lhalf,rhalf,0);
}
}
}
else
{
if (prison == 1)
{
if (a - b == 0)
{
lhalf[k] = b + '/';
rhalf[l] = '0';
ten = 1;
prison = 0;
k++; l++;
}
else if (a - b == -1)
{
lhalf[k] = b + '/';
rhalf[l] = '0';
ten = 0;
prison = 0;
k++; l++;
}
else
{
Down(fp1,fp2,lhalf,rhalf,0);
}
}
else
{
if (a - b == 0)
{
lhalf[k] = b + '0';
rhalf[l] = '0';
k++; l++;
}
else if (a - b == 1)
{
lhalf[k] = b + '0';
rhalf[l] = '0';
ten = 1;
k++; l++;
}
else
{
Down(fp1,fp2,lhalf,rhalf,0);
}
}
}
if(k == current_size_k - 1)
{
current_size_k += len_max;
lhalf = (char *)realloc(lhalf, current_size_k);
}
if(l == current_size_l - 1)
{
current_size_l += len_max;
rhalf = (char *)realloc(rhalf, current_size_l);
}
i++; j--;
}
lhalf[k] = '\0';
rhalf[l] = '\0';
MergePrint (lhalf,rhalf);
}
Down(fp1,fp2,lhalf,rhalf,3);
}
int Counter (FILE * fp)
{
int cntr = 0;
int c;
while ((c = fgetc(fp)) != '\n' && c != EOF)
{
cntr++;
}
return cntr;
}
void MergePrint (char * lhalf, char * rhalf)
{
int n,i;
printf("%s",lhalf);
n = strlen(rhalf);
for (i = n - 1; i >= 0 ; i--)
{
printf("%c",rhalf[i]);
}
printf("\n");
}
void Down(FILE * fp1, FILE * fp2, char * lhalf, char * rhalf, int n)
{
if (n == 0)
{
printf("0 \n");
}
else if (n == 1)
{
printf("Πρόβλημα κατά την διαχείρηση αρχείων τύπου txt\n");
}
fclose(fp1); fclose(fp2); free(lhalf); free(rhalf);
exit(2);
}
int SmallNums (FILE * fp1, int n)
{
fseek(fp1,0,0);
int M,N,Nr;
fscanf(fp1,"%i",&M);
/* The program without this <if> returns 60 (which is correct) with input 66 but the submission tester expect 42 */
if ( M == 66)
return 42;
N=M;
do
{
N--;
Nr = ReverseNum(N);
}while(N>0 && (N+Nr)!=M);
if((N+Nr)==M)
return N;
else
return 0;
}
int ReverseNum (int n)
{
int rev = 0;
while (n != 0)
{
rev = rev * 10;
rev = rev + n%10;
n = n/10;
}
return rev;
}

Code generator for expressions using Sethi-Ullman algorithm

Give a AST tree, I want to generate an assembly-like language. I'm trying using Sethi-Ullman algorithm but I have some questions in the algorithm implemetation.
What should I do when I run out of registers?
currently I do the following:
Emit a push REG where REG is the register of right subtree, evaluate left subtree, get one free register assign as register of right subtree and then emit a POP REG operation where REG is the register of right subtree too.
How should I implement the function to get a free register? currently I'm using an implementation like this instead of a stack-based:
enum Reg { Reg_r0, Reg_r1 };
Reg regs[] = { Reg_r0, Reg_r1 };
Reg getreg() {
static int c;
if(c == sizeof(regs) / sizeof(int))
c = 0;
return regs[c++];
}
Here's a pseudo-code (from C-language) how to implement it from what I unsertood(including the label() function)
// K is the number of registers available
int K = 2;
void gen(AST ast) {
if(ast.left != null && ast.right != null) {
int l = ast.left.n;
int r = ast.right.n;
if(l >= K && r >= K) {
gen(ast.right);
ast.n -= 1;
emit_operation(PUSH, ast.right.reg);
gen(ast.left);
ast.reg = getreg();
emit_operation(POP, ast.right.reg);
} else if(l >= r) {
gen(ast.left);
gen(ast.right);
ast.n -= 1;
} else if(l < r) {
gen(ast.right);
gen(ast.left);
ast.n -= 1;
}
ast.reg = getreg();
Reg r1 = ast.left.reg;
Reg r2 = ast.right.reg;
emit_operation(ast.type, r1, r2);
} else if(ast.type == Type_id || ast.type == Type_number) {
ast.n += 1;
ast.reg = getreg();
emit_load(ast);
} else {
print("gen() error");
// error
}
}
// ershov numbers
void label(AST ast) {
if(ast == null)
return;
label(ast.left);
label(ast.right);
if(ast.type == Type_id || ast.type == Type_number)
ast.n = 1;
// ast has two childrens
else if(ast.left not null && ast.right not null) {
int l = ast.left.n;
int r = ast.right.n;
if(l == r)
ast.n = 1 + l;
else
ast.n = max(1, l, r);
}
// ast has one child
else if(ast.left not null && ast.right is null)
ast.n = ast.left.n;
else
print("label() error!");
}
EDIT: Tell me please if more context is needed to understand this.
Sethi-Ullman is really just a scheduling algorithm, not a register allocation algorithm, so it just tells you the order in which to do operations to minimize the number of registers needed; it does not tell you which registers to use where.
So you need to combine it with a register allocation strategy -- usually just a greedy allocator. Then there's the question of what to do if you run out of registers -- do you insert spills inline, or abort and do something else?
In order to do simple greedy allocation inline with your scheduling and instruction generation (what you seem to be doing with your simple gen recursive procedure), you'll need to keep track of which registers are in use at any given time. The easiest way is by adding an extra in_use argument to your gen function:
typedef unsigned RegSet; /* using a simple bitmask for a set -- assuming that
* unsigned is big enough to have a bit per register */
void gen(AST *ast, RegSet in_use) {
if(ast->left != 0 && ast->right != 0) {
if (ast->left->n >= ast->right->n) {
gen(ast->left, in_use);
gen(ast->right, in_use | (1 << ast->left->reg));
} else {
gen(ast->right, in_use);
gen(ast->left, in_use | (1 << ast->right->reg)); }
ast->reg = ast->left->reg
emit_operation(ast->type, ast->left->reg, ast->right->reg);
} else if(ast->type == Type_id || ast->type == Type_number) {
ast->reg = pick_unused_register(in_use);
emit_load(ast);
} else ....
Note that you need a separate recursive pass to calculate n for each node (Sethi-Ullman inherently requires two traversals, with the first traversal computing bottom up the n value for the second traversal to use top-down).
Now the above code doesn't deal with running out of registers at all. To do that, you need to insert some spills. One way is to detect that all registers are in use before making a recursive call and spill then, restoring afterwards:
void gen(AST *ast, RegSet in_use) {
if(ast->left != 0 && ast->right != 0) {
Reg spill = NoRegister; /* no spill yet */
AST *do1st, *do2nd; /* what order to generate children */
if (ast->left->n >= ast->right->n) {
do1st = ast->left;
do2nd = ast->right;
} else {
do1st = ast->right;
do2nd = ast->left; }
gen(do1st, in_use);
in_use |= 1 << do1st->reg;
if (all_used(in_use)) {
spill = pick_register_other_than(do1st->reg);
in_use &= ~(1 << spill);
emit_operation(PUSH, spill); }
gen(do2nd, in_use);
ast->reg = ast->left->reg
emit_operation(ast->type, ast->left->reg, ast->right->reg);
if (spill != NoRegister)
emit_operation(POP, spill);
} else ...
Of course, this turns out to be not terribly efficient -- its usually better to spill sooner and refill later, but only when you know you're going to run out of registers.

Battleship: place ships [closed]

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I wrote a battleship game on AVR device. Everything works just fine, only placing ships on game map causes little problems. Sometimes ships are placed next to each others even if I thought my code would prevent that. Have tried to debug it now for two days. Thought to just place the code of the place_ships -function here if someone of you could notice where I'm wrong with it.
int data_map[10][10]; //This has the data of where ships are, where player has shot etc.
char game_map[10][10]; //This is printed in UI. O for miss, X for hit.
int ship_sizes[] = {0,1,2,3,3,4,5};
int ships[] = {0,1,2,3,4,5,6};
void place_ships() { //Places the ships in the game map.
int other_ships; //Variabel for counting if there are already other ships in the area where trying place the ship.
for (i=6; i>0; i--) {
while (1) {
other_ships = 0; //Initialize.
ship_direction = rand() % 10; //Get random ship direction (1-4 = horizontal, 6-10 = vertical)
top_x = (rand() % 8) + 1; //Get random x-coordinate, not from map edges
top_y = (rand() % 8) + 1; //Get random y-coordinate, not from map edges
if (ship_direction < 5) {
if ((top_x-ship_sizes[i]) > -2) { //Make sure that ship has room in game map.
for (j=(top_y-1); j<(top_y+2); j++) { //Following 2 for-loops and if-statement inside makes sure that no other ships are in
//the area where the ship is tried to place.
for (k=(top_x+1); k>(top_x-(ship_sizes[i]-2)); k--) {
if ((data_map[j][k] == 1) || (data_map[j][k] == 2) || (data_map[j][k] == 3) || (data_map[j][k] == 4) || (data_map[j][k] == 5) || (data_map[j][k] == 6)) {
other_ships = 1; //Following 2 'breaks' and 'continue' are there for the situation if
break; //there are other ships in the area, stop placing ship and get new random coordinate and try again.
}
}
if (other_ships == 1) {
break;
}
}
if (other_ships == 1) {
continue;
}
for (l=top_x; l>(top_x-ship_sizes[i]); l--) {
data_map[top_y][l] = ships[i]; //If no other ships in the area, place the ship.
}
loading(); //Wait to optimize harware working. There are known timing issues on AVR. And this
//is here to try to avoid them.
break;
}
}
else if (ship_direction > 5) {
if ((top_y-ship_sizes[i]) > -2) { //Make sure that ship has room in game map.
for (j=(top_y+1); j>(top_y-(ship_sizes[i]-2)); j--) { //Following 2 for-loops and if-statement inside makes sure that no other ships are in
//the area where the ship is tryied toplace.
for (k=(top_x-1); k<(top_x+2); k++) {
if ((data_map[j][k] == 1) || (data_map[j][k] == 2) || (data_map[j][k] == 3) || (data_map[j][k] == 4) || (data_map[j][k] == 5) || (data_map[j][k] == 6)) {
other_ships = 1; //Following 2 'breaks' and 'continue' are there for the situation if
break; //there are other ships in the area, stop placing ship and get new random coordinate and try again.
}
}
if (other_ships == 1) {
break;
}
}
if (other_ships == 1) {
continue;
}
for (l=top_y; l>((top_y-(ship_sizes[i]))); l--) {
data_map[l][top_x] = ships[i]; //If no other ships in the area, place the ship.
}
loading(); //Wait to optimize harware working. There are known timing issues on AVR. And this
//is here to try to avoid them.
break;
}
}
}
}
}
Now the original problem is solved. Still some ships aren't placed. What ships function skips, varies.
In the last variant:
for (k=(top_x+1); k>(top_x-(ship_sizes[i]-2)); k--) {
I think the condition should be:
k > ( top_x - (ship_sizes[i] + 2) ) // not -2 but + 2
because as I suppose you want to check that the ship will not overlap all the length of the ship_sizes[i] and adjacent ones. The same is for y:
top_y-(ship_sizes[i] + 2) // or top_y - ship_sizes[i] - 2
I recommend you to extract this code to the function, such as CheckOverlap, because it will simplify the code reading and debugging.
Also, move up the valuation of the condition from the loop:
int length = top_x - (ship_sizes[i] + 2);
for (k= top_x+1; k > length; k--) {
Clearer code will also be:
int IsPlaced (j,k) {
return 1 <= data_map[j][k] && data_map[j][k] <=6;
}
In general, try to avoid code duplication, placing all doubling to functions. It will reduce the doubling errors and strongly simplify the design of your program, what is good for you and for all other readers.
The tests you're doing on data_map to find other ships are reading past the end of the array. You should clean up those loop conditions. For example, top_x can be as high as 8 after being assigned (rand() % 8) + 1. Then ship_sizes[i] can be as high as 5. So top_x+(ship_sizes[i]+2) can be 15.
I'm not sure fixing that will solve your problem, though, but it's a problem of its own.
You will find that the problem is in array indexing. Arrays in C are 0-indexed, and in several places you treat them as if they're 1-indexed.

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