scanf ends my program and no error is thrown

Question

I am a beginner in C and I encountered problems when dealing with scanf, hoping for an explanation behind this

this is my code

#include <stdio.h>
#include <math.h>
#include <string.h>

int fav_charac() {
    char *string = ""; 
    printf("whats your fav letter?: ");
    scanf("%s", string);                         // the problem
    printf("your fav letter is %s\n", string);
}

int main() {
    fav_charac();
    return 0;
}

and this is the program running

whats your fav letter?: a

C:\Users\DELL\source\repos\hello C\x64\Debug\hello C.exe (process 15052) exited with code -1073741819.
Press any key to close this window . . .

as you can see it exited right after an input is made, why?

Does this answer your question? [How to prevent scanf causing a buffer overflow in C?](https://stackoverflow.com/questions/1621394/how-to-prevent-scanf-causing-a-buffer-overflow-in-c) — Karl Knechtel, Sep 03 '22 at 09:45

Vlad from Moscow · Answer 1 · 2022-04-03T11:41:17.687

In this declaration

char *string = "";

there is declared a pointer to an empty string literal that is represented in memory as one byte with the terminating zero value { '\0' }.

And in this call

scanf("%s", string);

you are trying to change the string literal pointed to by the pointer string. Any attempt to change a string literal results in undefined behavior.

From the C Standard (6.4.5 String literals)

7 It is unspecified whether these arrays are distinct provided their elements have the appropriate values. If the program attempts to modify such an array, the behavior is undefined.

By this reason it is always better to declare pointers to string literals with the qualifier const.

const char *string = "";

By the way in C++ opposite to C string literals have types of constant character arrays. So for such a declaration in C++

char *string = "";

the C++ compiler will issue a message.

If you want to enter more than one character then declare a character array as for example

char string[10] = "";

and use the following call of scanf

scanf( "%9s", string );

If you want to enter only one character then declare an object of the type char as for example

char c = '\0';

and use the following call of scanf

scanf( " %c", &c );

cediwelli · Accepted Answer · 2022-04-03T11:21:34.277

1

The scanf will attempt to write to string, which is .rodata. That means, the Data in string can only be read from but not written to. Use char string[64]; or something similar. This will place the variable into .data. Also limit user input in scanf with scanf("%63s", string); so that scanf won't write out of bounds.

Full example:

char string[64];
scanf("%63s", string);

edited Apr 03 '22 at 11:21

answered Apr 03 '22 at 10:46

cediwelli

370
1
8

thank you so much for the explanation, I thought `char *something` is a good practice thanks for correcting me. the code worked too. so `rodata` is immutable and `data` is mutable, from what I learnt `const` also creates immutable data, does this mean `*something` functions same as `const`? – Electron X Apr 03 '22 at 10:58
1

No its specific to strings. If you are initilizing a variable with a string-value, the string will be immutable. `char * str = "some_string";` and `char str2[] = "another_string";` will both be immutable. This is just how the compiler does things. The * means it is a pointer to a memory location. (arrays and pointers work pretty much the same that is why str[] and *str are basically interchangable). The important thing is the `[length]` which will cause the compiler to make the data mutable. – cediwelli Apr 03 '22 at 11:02
1

I recommend you look a bit into pointers, they are confusing but very powerful. When you understand the inner functionality of pointers, you wont think in data-types anymore but rather in bytes and this is pretty fun. – cediwelli Apr 03 '22 at 11:04
Off-by-one "`char string[64];` ... `scanf("%64s", string);`" should be `scanf("%63s", string);`. – chux - Reinstate Monica Apr 03 '22 at 11:14
@Cediwelli: Your 2nd example `char str2[] = "another_string";`, actually is mutable. It's initialized to `another_string`, but within its bounds can be altered. If you want `str2` to be immutable, you've to write `char const str2[] = …` (or `const char …`). – datenwolf Apr 03 '22 at 11:14
@chux-ReinstateMonica True. I just copy-pasted the 64, will edit. – cediwelli Apr 03 '22 at 11:17
@datenwolf Where in the obj is it stored? I just looked at the .data section and couldn't find it, which is why I assumed it is in .rodata but it is in neither of them? – cediwelli Apr 03 '22 at 11:23
@Cediwelli: You'd see it in `.data` only if it were in global scope. For variable in automatic storage the actual "memory" has no correspondence in the symbol table, the storage space is allocated at runtime (most usually from the stack) and initialized in "some" way (which can be anything but most usually is a copy from data in `.rodata`). – datenwolf Apr 03 '22 at 11:37

score 1 · Answer 3 · answered Apr 03 '22 at 10:46

1

try this:


 char string[50];  // [ the max of letters you want in your string ] 
    printf("whats your fav letter?: ");
    scanf("%s", string);

answered Apr 03 '22 at 10:46

Abdelmoula Bilel

71
8

`scanf("%s", string);` lacks a _width_ limit and is worse than `gets()`. – chux - Reinstate Monica Apr 03 '22 at 11:15

scanf ends my program and no error is thrown

3 Answers3