Why does scanf read every letter as 39?

Question

I used scanf("%d" ,&a), but for sake of curiosity I entered alphabets in the input console and to my surprise every time output was 39, why is it so?

I'm a newbie in language, so I made this program which reads/scans the input from the user and prints the output(sort of copy paste program) to get comfortable I used an integer as my variable value, but while executing the code, instead of entering integer I put letters [ex:- 'a', 'aaaaaaa', 'afadfad', 'awsw121', 'a1'] but every time the output was 39.

int a;
 printf("what`s your age in years?\n");

 scanf("%d" ,&a); 

 printf("your age is %d \n" , a);

whenever I put an input like ['1a' , '2afasfadfa' , '5awwee']

{ all input starting with an integer}

the answer comes out to be correct. I think that the code only reads the integer and as soon as it gets its input, It breaks the execution there itself.

Please look at this https://stackoverflow.com/help/how-to-ask — E.Praneeth, Jul 12 '19 at 04:45
In ASCII 39 stands for `'`. Are you putting characters inside single quotes? — niyasc, Jul 12 '19 at 04:48
When taken input it not valid than it try to display the garbage value. — Shivani Sonagara, Jul 12 '19 at 04:50
Did you expect any particular result? If not, you probanly want to change your approach to learning C. "Let's do weird things and see what happens" doesn't work well for this particular language. — n. m. could be an AI, Jul 12 '19 at 04:51
@Shivani Patel but why it doesn't print 39 in the latter part where I put input like [123abc] — shubham ranjan, Jul 12 '19 at 04:54
The ASCII value of `'` is irrelevant. You're printing an uninitialized variable. Never ignore the return value of `scanf`; it tells you how many items were successfully read: `if (scanf("%d", &a) != 1) { error }` — melpomene, Jul 12 '19 at 04:56
39 is just the value that `a` starts with because it is uninitialized. When you enter something that isn't a digit, no conversion is done and no value is stored in `a`. Check the return value from `scanf`. If what you type starts with a digit, this will be processed and converted until a non digit character is found. — 1201ProgramAlarm, Jul 12 '19 at 04:57
What did `scanf` return? (Hint: it probably wasn't 1, indicating 1 successful match.) — Steve Summit, Jul 12 '19 at 05:01
@shubhamranjan No, I don't mean the value of `a`, I mean the return value of the `scanf` function itself, if you do `r = scanf(...)`. — Steve Summit, Jul 12 '19 at 05:04
@Steve Summit scanf returns 0 when my input is a letter and it returns 1 when input is an integer — shubham ranjan, Jul 12 '19 at 05:14
Good. That's what it should do, and that's why you should check for it (as the answer you accepted also suggested). — Steve Summit, Jul 12 '19 at 05:15

Rishikesh Raje · Accepted Answer · 2019-07-12T06:24:26.913

When you have %d in scanf and you type a character, the scanf does not read anything. It actually returns a value of 0 elements read.

The value of 39 that you are getting is probably because you have declared int a on the stack and this will have any random value. Running the program again may or may not change this memory.

You can check this by initializing a with some other value e.g. int a = 5 and the output will always be 5.

Also you should always check the return value of scanf to see if you have actually read what you expect.

int a;
printf("what`s your age in years?\n");
if (scanf("%d" ,&a) != 1)
{
   // handle error
}   
printf("your age is %d \n" , a);

I think that the code only reads the integer and as soon as it gets its input, It breaks the execution there itself.

This is not entirely correct. scanf with %d will read the integer and the rest of the characters will remain in the buffer.

So, if you try another scanf with %d it will fail again.

Solution for this scenario is to read the buffer completely and then discard it. e.g. with

int c;
... 
while ((c = getchar()) != '\n' && c != EOF);

`while (scanf("%c",&dummy)==1);` will stop only on `EOF`. You probably want `int c; while ((c = getchar()) != '\n' && c != EOF);` — Spikatrix, Jul 12 '19 at 06:20

score 0 · Answer 2 · answered Jul 12 '19 at 05:03

scanf() returns the number of successful conversions. If the user types a number, it is converted and stored into variable and scanf() returns 1, if the user enters something that is not a number, such as AA, scanf() leaves the offending input in the standard input buffer and returns 0, finally if the end of file is reached (the user types ^Z enter in windows or ^D in unix), scanf() returns EOF

Why does scanf read every letter as 39?

2 Answers2