Help, I dont know why this wont work on linux, works on windows :(!

[tmand6]

#include<string.h>  
#include<stdio.h>  

    int main()  

    {  
     int i,j,n;  
     char s[51][100],ch1[200],ch2[200],temp[100];  
     while (scanf("%d",&n),n)  
     {  
      for (i=1;i<=n;i++)  
          scanf("%s",&s[i]);  
      for (i=1;i<n;i++)  
      for (j=i+1;j<=n;j++)  
      {  
       strcpy(ch1,s[i]);  
       strcpy(ch2,s[j]);  
       strcat(ch1,s[j]);  
       strcat(ch2,s[i]);  
       if (strcmp(ch1,ch2)<0) {strcpy(temp,s[i]); strcpy(s[i],s[j]); strcpy(s[j],temp);}  
      }  
      for (i=1;i<=n;i++)  
      printf("%s",s[i]);  
      printf("\n");  
     }  
     return 0;  
    }

works on Windows, not on Linux pls help

[jimblumberg]

You may want to increase your compiler warning level.

||=In function ‘main’:|
main.c|12|warning: format ‘%s’ expects argument of type ‘char *’, but argument 2 has type ‘char (*)[100]’ [-Wformat]|
||=== Build finished: 0 errors, 1 warnings ===|

You don't need the ampersand with C-strings.

Also remember that arrays in C/c++ start at zero and stop at size - 1.

Jim

[tmand6]

ty u saved me !

[rcgldr]

works on Windows, not on Linux.

This is the problem:

          scanf("%s",&s[i]);

If a program declares an array, like char array[...], then microsoft compilers will treat &array and array as essentially the same thing, but other compilers won't. As jimblumberg posted, you need to change this to:

          scanf("%s",s[i]);