I was writing a simple C program to take inputs from user for different variables using scanf() as follows:
#include <stdio.h>
int main(){
int a, b;
scanf("%d %d", &a, &b);
printf("%d\n",a);
printf("%d\n",b);
return 0;
}
The output for this is coming completely fine, as expected:
input: 10 23
output: 10
23
But then I also tried to take input for a string array, as follows(here char c[2] is a string array):
#include <stdio.h>
int main(){
int a, b;
char c[2];
scanf("%d %d %s", &a, &b, c);
printf("%d\n",a);
printf("%d\n",b);
printf("%s\n",c);
return 0;
}
And now the output was something unexpected:
input: 10 23 AM
output: 10
0
AM
Here, as it can be seen that the value being printed for variable b is coming to be 0, instead of the expected 23. How did taking input for a string array change the value of variable b?
Can someone help figure out, what wrong(of course silly mistake) have I done?emphasized text
Your char c[2]; variable is not a large enough array to hold a 2-character string plus the required nul-terminator. If you know that the input will be no more than 2 characters, changing to char c[3]; should be sufficient; however, for added safety, you can include a limit on how many characters will be read by the scanf call, using a format specifier like %2s (or, more generally, for an array such as char buff[n]; use a value of n - 1 in that format specifier).
As it stands, you have undefined behaviour – and this may include overwriting the value given for b (with the zero that should be the nul-terminator).
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