George2, perhaps you'd like to combine replies to several people in a single post
Any decent debugger will do. If you use MSVC, the built-in debugger could be used; if you use gcc then try gdb. There are also high-end commercial debuggers avaliable. A quick read of the debugger's documentation should give you a basic idea of how to use it for your program.
Code:#include <stdio.h> void J(char*a){int f,i=0,c='1';for(;a[i]!='0';++i)if(i==81){ puts(a);return;}for(;c<='9';++c){for(f=0;f<9;++f)if(a[i-i%27+i%9 /3*3+f/3*9+f%3]==c||a[i%9+f*9]==c||a[i-i%9+f]==c)goto e;a[i]=c;J(a);a[i] ='0';e:;}}int main(int c,char**v){int t=0;if(c>1){for(;v[1][ t];++t);if(t==81){J(v[1]);return 0;}}puts("sudoku [0-9]{81}");return 1;}
Perhaps for each function you call you could save the function name and arguments inside a global linked list, and then whenever you need to, print out out the contents of the linked list. Each time you exit a function, you can remove the last element of the linked list. A stack would work for this, but I think you want to traverse the structure without removing elements.Originally Posted by George2
Google cache: http://64.233.187.104/search?q=cache...s&ct=clnk&cd=2
I don't know man. Here's a bunch of bibliographic information about the article.
Code:@Article{Pescio:1997:STA, author = "Carlo Pescio", title = "Stack Trace Assertions Using {COFF}", journal = j-CCCUJ, volume = "15", number = "6", pages = "41--??", month = jun, year = "1997", CODEN = "CCUJEX", ISSN = "1075-2838", bibdate = "Fri May 16 08:01:41 1997", bibsource = "http://www.cuj.com/1506toc.htm", acknowledgement = ack-nhfb, }
Woah, easy there. A stack in terms of a dynamic array would do just fine, and be faster to boot. Better cache, less space, fewer instructions (push = increment counter + assign, pop = decrement counter).Perhaps for each function you call you could save the function name and arguments inside a global linked list
Code:#include <stdio.h> void J(char*a){int f,i=0,c='1';for(;a[i]!='0';++i)if(i==81){ puts(a);return;}for(;c<='9';++c){for(f=0;f<9;++f)if(a[i-i%27+i%9 /3*3+f/3*9+f%3]==c||a[i%9+f*9]==c||a[i-i%9+f]==c)goto e;a[i]=c;J(a);a[i] ='0';e:;}}int main(int c,char**v){int t=0;if(c>1){for(;v[1][ t];++t);if(t==81){J(v[1]);return 0;}}puts("sudoku [0-9]{81}");return 1;}
