Thread: escape sequence new line /n question

  1. #1
    Registered User
    Join Date
    Jan 2011
    Posts
    144

    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:
    #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);
    }

  2. #2
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    Join Date
    Feb 2012
    Posts
    347
    Quote Originally Posted by bos1234 View Post
    In the below code, why is it that if /n
    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]);    
        }
    }
    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

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