what will be the meaning of p in the following declarations?
458776
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
359046
#include <ctype.h>\n#include <stdio.h>\n#include <stdlib.h>\n\nstatic char *article[] = { "the","a", "one", "some", "any" };\nstatic char *noun   [] = { "boy","girl", "dog", "town", "car" };\nstatic char *verb   [] = { "drove","jumped", "ran", "walked", "skipped" };\nstatic char *prep   [] = { "to", "from", "over", "under", "on" };\n\nint\nmain()\n{\n  int asize = sizeof(article)/sizeof(article[0]);\n  int nsize = sizeof(noun)/sizeof(noun[0]);\n  int vsize = sizeof(verb)/sizeof(verb[0]);\n  int psize = sizeof(prep)/sizeof(prep[0]);\n\n  char buff[100];\n  for (int i=0; i<20; i++) {\n    (void)sprintf(buff, "%s %s %s %s %s %s.",\n      article[rand()%asize],\n      noun[rand()%nsize],\n      verb[rand()%vsize],\n      prep[rand()%psize],\n      article[rand()%asize],\n      noun[rand()%nsize]);\n    *buff = toupper(*buff);\n    (void)printf("%s&#92;n", buff);\n  }\n  return 0;\n}\n\n\n$ a.out\nSome town walked to a dog.\nThe car ran from any dog.\nAny boy drove under any town.\nOne girl jumped under the girl.\nOne car jumped to one girl.\nAny boy skipped over a dog.\nAny town jumped from some town.\nThe town ran from the town.\nOne car walked to any town.\nOne town ran over one boy.\nAny town jumped on the car.\nSome dog skipped from a boy.\nSome girl skipped over a boy.\nA car walked over some girl.\nThe car jumped from a girl.\nThe boy skipped on the town.\nAny car walked from one girl.\nAny dog ran to a town.\nA car jumped under some car.\nSome boy walked to the car.
803485
I think it's p=problem, i=infere' g=gather info? t=test e=expirement\n           good luck
208594
it means that there is a method or function that includes a parameter (or argument) called Start.  the number you pass into that method or function must be greater than zero.\n\nmyMethod(int Start) {\n...\n}\n\n//now to call method do this\nmyMethod(1);\n\n//but don't do this\nmyMethod(0);
413453
(P)laster (O)f (P)aris\n\nPOP!
1113307
a perfect day defines as:\nadd this up\n P+E+R+F+E+C+T = PERFECT  in all respect
710672
It means that you distribute (multiply) what is outside of the parentheses to each term within the parens..\n\nAre you finished with the little game you're playing, by the way?
367692
If p(1) is true and, for each integer k>1 the statements  p(1),p(2),....p(k) imply the statement p(k+1) then p(n) is true for all.\n\nIn other words if a statement is true for the integers 1..k\nand you can prove it from those for k+1, than it is true for all integers.\n\nIt is a stronger version of normal mathematical induction where one use P(k) to prove P(k+1).
143362
Yes, Zaphod is correct.\n\nYou should write, ATP and ADP - all capital letters.\nThe (P)i is sometimes written just as Pi - the letter'i' is a subscript. In this context the (P)i indicates an inorganic phosphate group.\n\nSo, adenosine TRIphosphate is hydrolysed into adenosine DIphosphate + an inorganic phosphate molecule. The interesting thing about this hydrolysis is that lots of energy is released when this particular bond is broken. This terminal phosphate bond is often described as a 'high-energy' bond.\nThe 'high-energy' bond is indicatd in diagrams as: Adenosine-P-P~P for ATP.
180273
d is two letters after b and b is capitalized.\n\nr is two letters after __ and __ is capitalized.\n\nThe answer is P.
869168
1.(c) 2.(d) 3.(b) 4.(a)
538319
A -> B\nB -> C\n...\nT -> U\nYou (U) -> V\nV -> W\n\nSo, You -> V
805034
#include<iostream.h>\nvoid main()\n{int a,b,c,d,e;\ncout<<"Enter any two numbers&#92;n";\ncin>>a>>b;\nc=a+b;\nd=a*b;\ne=a/b;\ncout<<"&#92;nThe sum of <<a<<|" and "<<b<<" is:"<<c;\ncout<<"&#92;nThe product of <<a<<|" and "<<b<<" is:"<<d;\ncout<<"&#92;nThe qoutient of <<a<<|" and "<<b<<" is:"<<e;\ngetch();\n}
741136
(a): Si+, P2+\n\n(b) Ar+, S-\n\n(c) Ca+, Cl2-\n\nBut the other answers that you gave are also correct.
1178626
:)D means mouth wide open smile and :)P means smile with tongue hanging out the side.\nAny good chatroom host can help you with anything related to the room or signs or abbreviations you do not understand. Such as LOL is laugh out loud and so on. Good luck!
222592
1. p'q\n\n2. p'\n\n3. pq' + p'\n\nI think i got the correct ones !
253055
Off the top of my head, don't hate me if there are errors...\n\nclass MyVar\n{\npublic:\n  int a;\n  MyVar ( MyVar& src )\n  { a = src.a; }\n}\n\n...\nMyVar p;\np.a = 5;\nMyVar q(p);\n// q.a==5 is true