escape sequence new line /n question

In the below code, why is it that if /n, it accepts 6 values, but if /n is not there, it accepts 5. The full code is below so that you may run the program

Code:

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#include <stdio.h>

int main()
{
        int temp[5]; //4 elements numbered 0 to 4
        int input;

        printf("How many degrees has it been the past 5 days?\n");
       
        for(input=0;input < 5;input++)
        {
                scanf("%f\n", &temp[input]);       
        }



        return(0);
}

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Quote:

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Originally Posted by bos1234

In the below code, why is it that if /n

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i am no expert in C but i found the following errors in your code. when you are saying /n i think you are actually referring \n

Code:

---

#include <stdio.h>

int main()
{
    int temp[5]; //4 elements numbered 0 to 4
 
  for(input=0;input < 5;input++)
    {
        scanf("%f\n", &temp[input]);   
    }
}

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the temp variable you are declaring as int but in scanf you are reading it as float, i think obviously you might have got warning in that and need to rectify it.

and the final part the scanf format specifier is of the following type
%[*][width][modifiers]type

scanf - C++ Reference
so it is improper if you put a \n in format specifier.

experts if i am wrong please correct me.

thanks and regards,
satya