int(*P(char *a))[10];\nint(*P(char *a[]));\nint(*P[10])(char a);\nint*(*P)(char *a[]);
207146
If you're asking what is the human explanation of\neach of these declarations, I would suggest turning\nto an old Unix utility which is freely available\non the net, "cdecl".\n\nIt's a little program which converts between\nC and C++ declaration and casting syntax and\na form of natural language.  \n\n"cdecl" explains these as follows:\n\n\nint(*P(char *a))[10];\ndeclare P as function (pointer to char) returning pointer to array 10 of int\n\nint(*P(char *a[]));\ndeclare P as function (array of pointer to char) returning pointer to int\n\nint(*P[10])(char a);\ndeclare P as array 10 of pointer to function (char) returning int\n\nint*(*P)(char *a[]);\ndeclare P as pointer to function (array of pointer to char) returning pointer to int
453545
here you go...\n(it'll work with up to 255 different disks, if you need more, you'll have to replace chars with ints)\n\nm1(char *a,b) {memove(b+1,b,strlen(b)+1); *b=*a; memmove(a+1,a, strlen(a);}\nm(char *a,b,c) {if (*a)  {m(a+1,b,c);m1(a,c);m(b,c,a);}
285035
c is a pointer to the 1st element in the array.\n\np and q are also pointers to the 1st element in the array.\n\nthe first loop prints *c which would be 2 (does this 5 times because it never changes the pointer c.\n\nthe second loop prints *p then increments p to the next element in the array (5 times)\n\nfull output would be:\n2222228344  (5 2s from the c pointer, 28344 from the p pointer)
208579
printf("%-15u&#92;n",40000);\n\nprovided I understood it correctly.
380360
I don't see how you could access elements in an array past MAXINT. (Using array syntax.)\n\nSo the practical answer is MAXINT.
299550
u didn't mentioned Language ...this is in C....\n\nint i, j, n;\nprintf("Enter no of lines");\nscanf ("%d",&n);\nfor (i=0;i<n;i++)\n{\n    for(j=0;j<=i;j++)\n    {\n        if(j==0||j==(i+1))\n          printf(" 1");\n        else\n          printf(" %d" ,i+2);\n     }\n     printf("&#92;n");\n}
224216
Delete (when used with a pointer) destroys what the pointer points to (e.g., pointer is still there, and can point to something else).  Destroy destroys the pointer/object itself.\n\nFor instance:\n\nchar** charArrayPtr = new char[50];\ndelete charArrayPtr;\ncharArrayPtr = NULL;\ncharArrayPtr = new char[25];\n\nBut if you did\n\ndestroy(charArrayPtr);\n\nyou couldn't use charArrayPtr anymore.\n\nReally, no real need to call destroy explicitly, for the most part.
559612
i don't remember the notation, but it is\n\nthe set of Integers not found on pages p sub 1 through p sub n
483613
char *Months[] = { "January", "February", "March", ... "December"};\n\nchar *GetMonth( int Index )\n{\n  return( Months[Index - 1] ); // Note -1 because first month is 1 not 0\n}
390615
Mn is close, but there are 2 errors:\n\n1) There is no len() function.  Use strlen() instead of len().\n\n2) He didn't allocate enough memory for the sprintf to fit the entire result string in the buffer.  The line needs room to store the &#92;n (or in windows, 2 chars - cr and lf ) in addition to the null terminator.  So, his code needs to add 2 instead of 1 to the length calculation.  \n\nTaking this a little further, the number of chars should be multiplied by the sizeof(t_char), which MIGHT NOT be 1 byte if the code is destined to use unicode.\n\nChances are, this is ascii, so the multiplication is unnecessary, but it is better to be safe than sorry.  Besides, if this was production code, you'd want to be able to use the same code in both places without any need to change it.  Nothing worse than having to go back and recalculate buffer manipulation code during a port.\n\nThe above subtleties demonstrate precisely how and why it is so very easy to make a mistake in c/c++, leading to a buffer underallocation or overwrite, and hence a security hole.
271646
I- + 2e- -------> I2\n\n4P + 3e-  -------------> P4\n\nthere ya go :)
405646
There are numerous ways to do that. \nHere is, probably the simplest:\n\nchar *s="char *s=%c%s%c;%cmain() {printf(s,34,s,34,10,10);}%c";\nmain() {printf(s,34,s,34,10,10);}\n\nThe program itself is fairly useless.\nThe whole point of an exercise like this is the satisfaction of finding a solution yourself. Too bad, you did not give yourself a chance
436488
for (int i = 0; i < arrayA.length; i++)\nfor (int j = 0; j < arrayA[0].length; j++)\n  arrayC[i][j] = arrayA[i][j] + arrayB[i][j];\n\nSystem.out.println("  Values for arrayC");\nfor (int i = 0; i < arrayC.length; i++)\n   for (int j = 0; j < arrayC[0].length; j++)\n   {\n      System.out.print(arrayC[i][j]);\n      if(j < arrayC[0].length - 1)\n        System.out.print("   ");\n      else\n        System.out.println("");\n  }
188665
#include <stdio.h>\n#include <stdlib.h>\n\nint\nmain(int argc, char* argv[])\n{\n  int count = argc>1 ? atoi(argv[1]) : 10;\n  FILE *f = fopen("numfile", "w+");\n  if (!f)\n    perror("fopen"), exit(-1);\n  for (int i=0; i<count; i++)\n    fprintf(f, "%d&#92;n", rand());\n  fclose(f);\n  return 0;\n}
182444
static_cast<int>("A") = 65\n\ngives you the ASCII value of the character\n\nWhole program:==============================================\n\nchar array[25]={'A','B',.....'Z'}\n\ncout << "Enter number: ";\ncin >> number\n\nif(number <= 10){\n  cout << number;\n}\nelse\n cout << array[number- 10];\n\nThat is the basic program you need to write it will work just dress it up to look pretty
263655
hmm.. let me walk you through the pseudocode\n\nint find_largest(int array[], int start, int finish) {\n   if (start==finish) \n       return start;\n   int middle = (start+finish)/2;\n   int max_1st_set = find_largest(array, start, middle);\n   int max_2nd_set = find_largest(array, middle+1, finish);\n   int max = (array[max_1st_set] > array[max_2nd_set])? max_1st_set:max_2nd_set;\n   return max;\n}\n\nSomething like this should work for you!\nI am not going to do your entire home work so I leave the rest of the analysis to you! :-)
2251
i dont know in what environment you work, but a C++ program would look like this:\n#include<iostream.h>\nint n,a[10][10],i,j;\nvoid main()\n{cin>>n;\n for (i=0;i<n;i++)\n   for (j=0;j<n;j++)\n      {\n       cin>>a[i][j];\n       if (a[i][j]==0) cout<<i<<" "<<j<<"&#92;n";\n      }\n}