Hexadecimal conversion decimal-CodePudding

/* programs fill in the blanks, don't change related to the input and output statements,
Input a number of consecutive # at the end of the string (the length of the string does not exceed 80), in case # # all input end,
Do the following with each string: filter after all the non hexadecimal character, form a new string (hexadecimal form), then it is converted to a decimal number after output,
Example: the brackets is
Input:
10 + 1 + # # # - # Pf4 (three consecutive input string)
Output:
16
3905
10 */
# include
# define MAXLEN 80
Int main ()
{int I, k;
Long number;
Char ch, oldch, STR [MAXLEN], num [MAXLEN];

Oldch=' ';
While ((ch=getchar ())!='#' | | oldch!='#')
{
k=0;
While (ch!='#' & amp; & k{
STR [k++]=ch;
ch=getchar();
}
Oldch='#';
STR [k]=0;
/* -- -- -- -- -- -- -- -- - */
Printf (" % ld \ n ", number).
}
}

CodePudding user response:

Printf (" % ld \ n ", number).
You are not the assignment, the number,,

CodePudding user response:

reference 1st floor fireyou response:

printf (" % ld \ n ", number).
You are not the assignment, the number,,

Is to fill in the blanks, this is about architecture

CodePudding user response:

 
# include 
# define MAXLEN 80 

Int main () 
{
Int I, k; 
Long number; 
Char ch, oldch, STR [MAXLEN], num [MAXLEN]; 

Oldch=' '; 
While ((ch=getchar ())!='#' | | oldch!='#') 
{
k=0; 
While (ch!='#' & amp; & k{
STR [k++]=ch; 
ch=getchar(); 
} 
Oldch='#'; 
STR [k]=0; 
/* -- -- -- -- -- -- -- -- - */
# define isDigit (x) (x) & gt;='0' & amp; & (x) & lt;='9') 
# define isA2F (x) (x) & gt;='A' & amp; & (x) & lt; ) 
='F'# define isa2f (x) (x) & gt;='a' & amp; & (x) & lt; ) 
='f'# define isXDigit (x) (isDigit (x) | | isA2F (x) | | isA2F (x)) 
Number=0; 
for (i=0; I & lt; k; I++) 
{
If (isXDigit (STR [I])) 
Num=STR [number++] [I]; 
} 
Num [number]=0; 
K=number; 
Number=0; 
for (i=0; I & lt; k; I++) 
{
If (isDigit (num [I])) 
Number=number * 16 + (num [I] - '0'); 
Else if (isA2F (num) [I]) 
Number 16 + 10=number * + (num [I] - 'A'); 
Else if (isa2f (num) [I]) 
Number 16 + 10=number * + (num [I] - 'a'); 
} 
Printf (" % ld \ n ", number). 
} 
}

CodePudding user response:

Thank you, although a little look not to understand, but still want to thank you

CodePudding user response:

Input a number of consecutive # by the end of the string ( the length of the string does not exceed 80 ), encounter # # all input end,

CodePudding user response:

I hope it can help you: https://blog.csdn.net/it_xiangqiang/category_10581430.html
I hope it can help you: https://blog.csdn.net/it_xiangqiang/category_10768339.html