How To Fix "return Value Ignored: 'scanf'" Code C6031 In Visual Studio

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Learn more about Labs How to fix "return value ignored: 'scanf'" code C6031 in visual studio Ask Question Asked 5 years, 2 months ago Modified 1 year ago Viewed 38k times 4

I am brand new to coding C (and coding in general) so I have been practicing with some random programs. This one is supposed to determine the cost of a transit ticket (Translink Vancouver prices) based on the user's age and the desired number of "zones" (how far they would like to go). I have compiled it successfully but for some reason which I can not figure out, the scanf functions are being ignored. How do I fix this? Please keep in mind I have only been coding for a few days. Thanks!

int main(void) { int zones; int age; double price = 0.00; printf("Welcome to TransLink cost calculator!\n\n"); printf("Please enter the desired number of zones (1, 2, or 3) you wish to travel: "); scanf("%d", &zones); if (zones < 1) { printf("Invalid entry\n"); price = 0.00; } else if (zones > 3) { printf("Invalid entry\n"); price = 0.00; } else if (zones == 1) { printf("Please enter your age: "); scanf("%d", &age); if (age < 0.00) { printf("Invalid Aage"); } else if (age < 5) { price = 1.95; } else if (age >= 5) { price = 3.00; } } else if (zones == 2) { printf("Please enter your age: "); scanf("%d", &age); if (age < 0) { printf("Invalid Aage"); } else if (age < 5) { price = 2.95; } else if (age >= 5) { price = 4.25; } } else if (zones == 3) { printf("Please enter your age: "); scanf("%d", &age); if (age < 0) { printf("Invalid Aage"); } else if (age < 5) { price = 3.95; } else if (age >= 5) { price = 4.75; } } printf("The price of your ticket is: $%.2f + tax\n", price); system("PAUSE"); return 0; }

• c

Share Improve this question Follow asked Sep 19, 2019 at 5:25 Caiden KellerCaiden Keller 4111 gold badge11 silver badge22 bronze badges 5

• 2 Please explain how you use the value returned by this call scanf("%d", &age);. Not the value scanned into age, the value returned by that call. The value which tells you whether scanning has succeeded. If you cannot, then you have the answer to your question. – Yunnosch Commented Sep 19, 2019 at 5:28
• Thanks for your response but I don't understand what you are trying to say. I am not sure why it is not working (hence my post here) so if you know how I can fix it and could tell me that would be great, thanks. – Caiden Keller Commented Sep 19, 2019 at 5:37
• btw. It's not the problem , but age is an int and your are comparing to a float constant: if (age < 0.00) { – Mike Commented Sep 19, 2019 at 5:41
• 1 scanf() returns the number of values processed. You should compare this to the number of variables you're trying to read, to make sure it was successful. e.g. if (scanf("%d", &age) != 1) { // print error message } – Barmar Commented Sep 19, 2019 at 6:03
• 1 Whatever you did to make visual studio complain about ignoring scanf()s return value. Please teach it to as many people as you can. – Yunnosch Commented Sep 19, 2019 at 6:14

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4 Answers 4

Sorted by: Reset to default Highest score (default) Trending (recent votes count more) Date modified (newest first) Date created (oldest first) 6

A bit too much here to put in a comment.

I use a version of Visual C but it never complains about the return value from scanf not being used. What it does is to complain that scanf is unsafe and deprecated, when it isn't.

MS thinks I should be using its own "safer" version scanf_s which is even tricker to use and IMO no safer at all – because it is not a like-for-like replacement but takes different arguments, and so it is easy to make mistakes in using it.

One consequent problem is the compiler issues a warning for every use of scanf (and some other functions) which obscures other warnings. I deal with it as advised by adding a #define before the first library header inclusion.

#define _CRT_SECURE_NO_WARNINGS #include <stdio.h>

There are other matters which MS warns about too, and I actually place three #defines at the start of each file:

#define _CRT_SECURE_NO_WARNINGS #define _CRT_SECURE_NO_DEPRECATE #define _CRT_NONSTDC_NO_DEPRECATE #include <stdio.h>

And now the relevant warnings are easy to see.

Share Improve this answer Follow answered Sep 19, 2019 at 7:42 Weather VaneWeather Vane 34.5k77 gold badges4141 silver badges6161 bronze badges 2

• I fully agree with the advice to disable MSVC's advice to use their pseudo-Annex K functions; however, the scanf family are unsafe, for several reasons, most importantly that %s is just as dangerous as gets, and that numeric conversion has undefined behavior on numeric input overflow (yes, that means the C library is allowed to crash just because someone typed too many digits). – zwol Commented Jun 9, 2021 at 15:08
• @zwol yes, but C is unsafe anyway. You can create a segfault very easily by exceeding array bounds. My point was that MS's scanf_s isn't really any better than scanf - both can cause damage when used incorrectly. – Weather Vane Commented Jun 9, 2021 at 18:33

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From documentation of scanf() (e.g. https://en.cppreference.com/w/c/io/fscanf)

Return value 1-3) Number of receiving arguments successfully assigned (which may be zero in case a matching failure occurred before the first receiving argument was assigned), or EOF if input failure occurs before the first receiving argument was assigned.

You are ignoring that return value.

Replace

scanf("%d", &age);

by

int NofScannedArguments=0; /* Number of arguments which were successfully filled by the most recent call to scanf() */ /* ... do above once, at the start of your function */ NofScannedArguments= scanf("%d", &age); /* check the return value to find out whether scanning was successful */ if(NofScannedArguments!=1) /* should be one number */ { exit(EXIT_FAILURE); /* failure, assumptions of program are not met */ }

... to find out whether scanning succeeded. Not doing so is a bad idea and worth the warning you got.

In case you want to handle failures more gracefully, e.g. prompt the user again, use a loop and read http://sekrit.de/webdocs/c/beginners-guide-away-from-scanf.html about the pitfalls you may encounter. I am not saying you should NOT use scanf, the article explains a lot about using scanf, while trying to convince you not to.

Share Improve this answer Follow edited Sep 19, 2019 at 6:21 answered Sep 19, 2019 at 6:05 YunnoschYunnosch 26.6k99 gold badges4444 silver badges6161 bronze badges 4

• If you'd used a more reasonable variable name than the essay you chose… I often use rv (return value) — I accept that's too minimal, but retval is a possibility, with an underscore in the middle if desired. – Jonathan Leffler Commented Sep 19, 2019 at 6:15
• 1 @JonathanLeffler I think I know what you mean. But in this case OP needs all the super-explicit explanation they can get. Let me see whether I can get both needs covered. There. Deal? – Yunnosch Commented Sep 19, 2019 at 6:18
• IMO this (making a variable) is not really useful as it really does nothing. in competitive programming at least – J Muzhen Commented Jan 6, 2022 at 13:03
• @JMuzhen I think I know what you mean. But in this case OP needs all the super-explicit explanation they can get. – Yunnosch Commented Jan 6, 2022 at 17:38

Add a comment | 0

I know this is an old question and asked several times in other threads.

To suppress individual C6031 warning, I cast the function w/ (void).

Example:

double x; (void)scanf("%lf", &x);

this way, I can "mark" that I'm intentionally ignoring the return value of the function (=scanf). This is useful for the functions such as InitializeCriticalSectionAndSpinCount(), which always succeeds and returns non-zero value, worth ignoring safely.

This seems better to suppress C6031 entirely.

Share Improve this answer Follow edited Nov 8, 2023 at 12:25 answered Nov 8, 2023 at 1:12 beshiobeshio 80422 gold badges88 silver badges1818 bronze badges Add a comment | -1

Using C++ functions for input is SO much easier. Instead of scanf and printf one could use cin and cout as the following demonstrates:

#include <iostream> // for cin and cout use int main() { int zones; std::cout << "Enter zones" << std::endl; // endl is similar to \n std::cin >> zones; std::cout << "Your zones is " << zones << std::endl; } Share Improve this answer Follow edited Jun 9, 2021 at 15:00 answered Jun 9, 2021 at 14:40 KenKen 1411 bronze badge 2

• Firstly, the author was clearly wanting to know how to use scanf and printf; secondly you are clearly not experienced with the scanf and printf functions, as although they may be harder to use, they run much faster than std::cin and std:cout. – J Muzhen Commented Jan 6, 2022 at 13:00
• This is not what author asked – Afzal Ali Commented Nov 17, 2022 at 8:50

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