Error Assignment Of Read-only Variable
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Assignment Of Read Only Variable C++
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Read Only Variable Not Assignable
assignment of read-only location up vote 0 down vote favorite When I compile this program, I keep getting this error example4.c: In function ‘h’: example4.c:36: error: assignment of read-only location example4.c:37: error: assignment of read-only location I think it has something to read only variable is not assignable xcode c++ do with the pointer. how do i go about fixing this. does it have to do with constant pointers being pointed to constant pointers? code #include #include #include "example4.h" int main() { Record value , *ptr; ptr = &value; value.x = 1; strcpy(value.s, "XYZ"); f(ptr); printf("\nValue of x %d", ptr -> x); printf("\nValue of s %s", ptr->s); return 0; } void f(Record *r) { r->x *= 10; (*r).s[0] = 'A'; } void g(Record r) { r.x *= 100; r.s[0] = 'B'; } error assignment of read only location void h(const Record r) { r.x *= 1000; r.s[0] = 'C'; } c compiler-errors share|improve this question asked Apr 25 '13 at 23:22 monkey doodle 3052714 1 void h(const Record r) :It has declared that it isn't supposed to change the r. –BLUEPIXY Apr 25 '13 at 23:28 Declaring a parameter as const doesn't do much; it merely prevents the function from modifying its local copy. That can be a good idea, since it documents and enforces your intent to leave it with the original value that was passed in, but it has no effect on the caller. –Keith Thompson Apr 25 '13 at 23:37 See A 'conflicting types' error for the sequel — not a duplicate, but closely related. –Jonathan Leffler Apr 25 '13 at 23:54 add a comment| 1 Answer 1 active oldest votes up vote 4 down vote accepted In your function h you have declared that r is a copy of a constant Record -- therefore, you cannot change r or any part of it -- it's constant. Apply the right-left rule in reading it. Note, too, that you are passing a copy of r to the function h() -- if you want to modify r then you must pass a non-constant pointer. void h( Record* r) { r->x *= 1000; r->s[0] = 'C'; } share|improve this answer edited Apr 25 '13 at 23:30 answered Apr 25 '13 at 23:24 K Scott Piel 3,576519 add a comment| Your Answer draft saved draft
am using Xcode on a mac. I get the following error messages when I try to run
Read Only Variable In Java
this... error: assignment of read-only variable 'federalWithholding' error: assignment of read-only variable read only variable javascript 'stateWithholding' error: assignment of read-only variable 'hospitalization' error: assignment of read-only variable 'unionDues' can anyone tell me
Assignment Of Read-only Location C
why this is happening? heres the program so far... #include #include #include const double federalWithholding = 0.18; const double stateWithholding = 0.045; const double hospitalization http://stackoverflow.com/questions/16226313/error-assignment-of-read-only-location = 25.65; const double unionDues = 0.02; using namespace std; int main () { string employeeInitials; double hoursWorked, hourlyRate, totalWages, totalDeductions, netPay; cout << "Enter employee initials: " << endl; getline (cin,employeeInitials); cout << "Enter Hours Worked" << endl; cin >> hoursWorked; cout << "Enter Hourly Rate" << endl; cin >> hourlyRate; cin >> totalWages = http://www.cplusplus.com/forum/beginner/37164/ hoursWorked * hourlyRate; federalWithholding = 0.18 * totalWages; stateWithholding = 0.045 * totalWages; hospitalization = 25.65; unionDues = 0.02 * totalWages; totalDeductions = federalWithholding + stateWithholding + hospitalization + unionDues; netPay = totalWages - totalDeductions; cout << fixed << showpoint << setprecision(2); cout << "Employee Initials: " << employeeInitials << endl; cout << "Hours Worked: " << hoursWorked << endl; cout << "Total Wages: " << totalWages << endl; cout << "\nDeductions"; cout << "Federal Withholding: " << federalWithholding << endl; cout << "State Withholding: " << stateWithholding << endl; cout << "Hospitalization: " << hospitalization << endl; cout << "Union Dues: " << unionDues << endl; cout << "\nTotal Deductions: " << totalDeductions << endl; cout << "Net Pay: " << netPay << endl; return 0; } Feb 26, 2011 at 9:18pm UTC Moschops (7069) const double federalWithholding = 0.18; federalWithholding = 0.18 * totalWages; Feb 26, 2011 at 9:19pm UTC firedraco (6199) All of the variables you are assigning to are const which means you can
question and get tips & solutions from a community of 418,504 IT Pros & Developers. It's quick & easy. assignment of read-only variable P: n/a yinglcs Hi, I have a method like this: const A* getMaxArea(const vector& aList); And in my caller, https://bytes.com/topic/c/answers/485598-assignment-read-only-variable i have this: A* const a = NULL; if (!aList.empty()) { a = getMaxArea(aList); //..... } http://forum.arduino.cc/index.php?topic=42406.0 But I get this compile error: .../NavBarGroupResultListProcessor.cpp:49: error: assignment of read-only variable 'a' Can you please tell me what is the pointer, i think A* const a // is a pointer to 'const a', not a const pointer to 'a', so I should be able to assign it, right? Apr 29 '06 #1 Post Reply Share this Question 5 Replies P: n/a Dennis Jones wrote read only in message news:11**********************@i39g2000cwa.googlegr oups.com... Hi, I have a method like this: const A* getMaxArea(const vector& aList); And in my caller, i have this: A* const a = NULL; if (!aList.empty()) { a = getMaxArea(aList); //..... } But I get this compile error: ../NavBarGroupResultListProcessor.cpp:49: error: assignment of read-only variable 'a' Can you please tell me what is the pointer, i think A* const a // is a pointer to 'const a', not a const pointer to 'a', so I should be able to assign read only variable it, right? I believe it should be read from right to left: 'a' is a const pointer to 'A'. Therefore, 'a' (the pointer) is constant and may not be re-assigned. - Dennis Apr 29 '06 #2 P: n/a Jakob Bieling yi*****@gmail.com wrote: I have a method like this: const A* getMaxArea(const vector& aList); And in my caller, i have this: A* const a = NULL; if (!aList.empty()) { a = getMaxArea(aList); //..... } But I get this compile error: ../NavBarGroupResultListProcessor.cpp:49: error: assignment of read-only variable 'a' Can you please tell me what is the pointer, i think A* const a // is a pointer to 'const a', not a const pointer to 'a', so I should be able to assign it, right? No, it is a const pointer to an object of type 'A'. Always read the types from right to left. Changing the type to 'A const*' will fix your problem, as far as I can see. hth -- jb (reply address in rot13, unscramble first) Apr 29 '06 #3 P: n/a yinglcs Thanks. It works. I have another question: I have another funcion, which is this: void f1(const A* aPtr); A const *a = NULL; if (!aList.empty()) { a = getMaxArea(aList); // try to call f1() here, const A* a1 = a; f1 (a1); // won't compile due to error: invalid conversion from 'const A*' to 'A*' } Any idea? Apr 29 '06 #4 P: n/a Andre Kostur On Sat, 29 Apr 2006 09:29
(read only) > Software > Syntax & Programs > a very basic question Print Go Down Pages: [1] Topic: a very basic question(Read 3521 times) previous topic - next topic grimmjaw Newbie Posts: 6 Karma: 0[add] Arduino rocks a very basic question Sep 19, 2009, 11:17 am i 'm trying to write a program that triger my food dispensing rig every four hours.but for the love of god ,the programming just keep on giving me error at the the first few line at #defineerror: 'voip' does not name a typethis is the program#define Rled 13#define Gled 12#define feeder 11int mins = 0;int hours = 0;voip setup(){ pinMODE (RLed,OUTPUT); pinMODE (Gled,OUTPUT); pinMODE (Feeder,OUTPUT);}voip loop() { digitalwrite(Rled ,HIGH); for (hours=0;hours <4 ;hours ++){ for (mins=0;mins < 59;mins ++){ delay (60000); } } digitalwrite (Gled,HIGH); digitalwrite (feeder,HIGH); delay (5000); digitalwrite(Gled,LOW); digitalwrite(Feeder,LOW); }i konw this may be a simple error.sorry for the noob question CaptainObvious God Member Posts: 686 Karma: 3[add] My name is Bob, and I'm an addict. Re: a very basic question #1 Sep 19, 2009, 11:21 am I'm not sure if this is the EXACT code you have.. but if so;Your SETUP and LOOP are both using.. voip? Should be void.As well as your pinMode.. it needs to be exact. C++ is case sensitive.Instead of pinMODE, use pinMode.And when you turn your feeder to output.. you use a captial F, which will cause errors.. it needs to be exact.pinMode(feeder, OUTPUT);Along with all the digitalWrites... they need to be digitalWrite.. not digitalWRITE, or digitalwrite, or DigitalWrite.. case sensitive!And again at the bottom, you use a capital F in feeder. Make sure everything is case sensitive grimmjaw Newbie Posts: 6 Karma: 0[add] Arduino rocks Re: a very basic question #2 Sep 19, 2009, 12:40 pm hahahah..i know that it would be something stupid like that. .thanks again CaptainObvious God Member Posts: 686 Karma: 3[add] My name is Bob, and I'm an addict. Re: a very basic question #3 Sep 19, 2009, 12:45 pm Trust me.. I spent almost 2 hours on a sketch just like that my first couple weeks! Haha, it just takes some getting used to!:PGood luck, glad to see you have a good attitude about it!:D ahdavidson Sr. Member Posts: 250 Karma: 6[add]