How do I subtract 1 from an array of integers - arrays

I am learning C in school, and I am having a little difficulty on my project.
Basically, I am writing a function that gets a string containing a positive integer. The function subtracts 1 from that integer and puts the obtained value in the string.
So , if I have this ;
char nums[] = "2462";
how do I write a function that will subtract 1 from the integer, so that the result is "2461"??

First, convert the character array into an integer.
You can use atoi (ASCII to Integer), but because it returns 0 on error there's no way to tell the difference between successfully converting "0" and an error.
Instead use strtol (STRing TO Long integer).
// end stores where parsing stopped.
char *end;
// Convert nums as a base 10 integer.
// Store where the parsing stopped in end.
long as_long = strtol(nums, &end, 10);
// If parsing failed, end will point to the start of the string.
if (nums == end) {
perror("Parsing nums failed");
}
Now you can subtract, turn the integer back into a string with sprintf, and put it in nums.
sprintf(nums, "%ld", as_long - 1);
This is not entirely safe. Consider if nums is "0". It only has 1 byte of space. If we subtract 1 then we have "-1" and we're storing 2 characters where we only have memory for 1.
For the full explanation of how to do this safely, see How to convert an int to string in C?.
Alternatively, don't store it, just print it.
printf("%ld", as_long - 1);

One way is to convert string -> int -> string.
You can do this by using atoi and sprintf.
Simple implementation (far from perfect):
#include <stdlib.h>
#include <stdio.h>
int main()
{
int a;
char b[5];
a = atoi("2462");
a--;
sprintf(b, "%d", a);
printf("%s\n", b);
return 1;
}

Related

Using sprintf to convert an unsigned long to char array producing 0

I am trying to convert an unsigned long using the sprintf function in c. Code goes like:
char ID[6];
sprintf(ID,"%lu",a.id);
a.id is a number that is passed in that can range from 0 > but I only want the first 6 regardless. Using printf("%lu",a.id); prior to conversion prints the right number but once I try print the string from the char the outcome is 0. Not too sure why this is happening any advice would be much appreciated.
... once I try print the string from the char the outcome is 0
Code is risking undefined behavior (UB): buffer overrun, potential wrong specifier.
but I only want the first 6 regardless.
Insure the buffer is big enough for 6 characters and a terminating null character.
//char ID[6];
char ID[6+1];
Handle a.id outside the expected range of 0...999999 with % some_unsigned_constant. This does print the last 6.
// sprintf(ID,"%lu",a.id);
sprintf(ID,"%lu",a.id % 1000000u);
As type of a.id, not posted, 2 steps may be useful to make sure a matching print specifier is used.
// sprintf(ID,"%lu",a.id % 1000000u);
unsigned long ul = a.id
sprintf(ID,"%lu", ul % 1000000u);
To print the first six, even if outside the 0...999999 range, use snprintf() which will print only up to the first 6.
char ID[6+1];
snprintf(ID, sizeof ID, "%lu", (unsigned long) a.id);
And the correct way to do what the OP wants is to
#include <stdio.h>
int main()
{
char ID[6];
unsigned long x = 3453342432;
int r = snprintf(ID, sizeof(ID), "%lu",x);
printf("%d %s\n", r, ID);
return 0;
}
Note that the snprintf result is a number of characters that would had been written. It was noted by others that you must reserve one character extra space for terminating NULL.

Converting a binary string to integer

When I try and convert my binary string to int I am receiving a couple of mistakes that I can not figure out. First I am reading from a file and the leading zeros are not showing up when I convert and the new line is showing zero.
This code I am using from this questions: Convert binary string to hexadecimal string C
char* binaryString[100];
// convert binary string to integer
int value = (int)strtol(binaryString, NULL, 2);
//output string as int
printf("%i \n",value)
My txt file and what I am expecting as an output:
00000000
000000010001001000111010
00000000000000000000000000000001
101010111100110100110001001001000101
What I get:
0
0
70202
1
-1127017915
This line:
char* binaryString[100];
Is declaring an array of 100 char pointers (or 100 strings). You probably meant this to declare a buffer of 100 characters to be interpreted as a single string:
char binaryString[100];
char *binaryString[100];
// You are creating an array of pointers in this scenario, use char binaryString[100] instead;
int value = (int)strtol(binaryString, NULL, 2);
// 101010111100110100110001001001000101 Is a 36 bit number, int (in most implementations) is 32 bit. use long long (64 bit in visual c++) as type and strtoll as function instead.
printf("%i \n",value)
Must be printf("%lld \n", value).
In summary:
#include "stdio.h"
#include "stdlib.h" // required for strtoll
int main(void)
{
char str[100] = "101010111100110100110001001001000101";
long long val = 0;
val = strtoll(str, NULL, 2);
//output string as int
printf("%lld \n", val);
return 0;
}
if im understanding this correctly you want to take a binary string so ones and zeros and convert it to a Hex string so 0-F, if so the problem is with the Write not the Convert, you specified '%i' as the written value format, what you need to do for hex is specify '%x'
Change this "printf("%i \n",value)" to "printf("%x\n",value)"

atoi ignores a letter in the string to convert

I'm using atoi to convert a string integer value into integer.
But first I wanted to test different cases of the function so I have used the following code
#include <stdio.h>
int main(void)
{
char *a ="01e";
char *b = "0e1";
char *c= "e01";
int e=0,f=0,g=0;
e=atoi(a);
f=atoi(b);
g=atoi(c);
printf("e= %d f= %d g=%d ",e,f,g);
return 0;
}
this code returns e= 1 f= 0 g=0
I don't get why it returns 1 for "01e"
that's because atoi is an unsafe and obsolete function to parse integers.
It parses & stops when a non-digit is encountered, even if the text is globally not a number.
If the first encountered char is not a space or a digit (or a plus/minus sign), it just returns 0
Good luck figuring out if user input is valid with those (at least scanf-type functions are able to return 0 or 1 whether the string cannot be parsed at all as an integer, even if they have the same behaviour with strings starting with integers) ...
It's safer to use functions such as strtol which checks that the whole string is a number, and are even able to tell you from which character it is invalid when parsing with the proper options set.
Example of usage:
const char *string_as_number = "01e";
char *temp;
long value = strtol(string_as_number,&temp,10); // using base 10
if (temp != string_as_number && *temp == '\0')
{
// okay, string is not empty (or not only spaces) & properly parsed till the end as an integer number: we can trust "value"
}
else
{
printf("Cannot parse string: junk chars found at %s\n",temp);
}
You are missing an opportunity: Write your own atoi. Call it Input2Integer or something other than atoi.
int Input2Integer( Str )
Note, you have a pointer to a string and you will need to establish when to start, how to calculate the result and when to end.
First: Set return value to zero.
Second: Loop over string while it is not null '\0'.
Third: return when the input character is not a valid digit.
Fourth: modify the return value based on the valid input character.
Then come back and explain why atoi works the way it does. You will learn. We will smile.

how to print a string in c without pā” in the end of the string

so i've been writing a program that convert a decimal number to it's boolean representation but every time i compile the return value which is a string show additional characters like pā” here is the program
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
void main (void)
{
signed char str[256];
int dec,rest;
int i = -1;
int j;
printf ("write a decimal number : ");
scanf ("%d",&dec);
do
{
rest = dec % 2;
dec/= 2;
for (j=i;j>-1;--j)
str[j+1]=str[j];
str[0]=rest+48;
++i;
} while (dec!=0);
printf ("the binary representation of this number is %s",str);
}
the output :
write a decimal number : 156
the binary representation of this number is 10011100pā”
i don't know if im missing something but i will be grateful if you guys help me
In C and C++, strings are null-terminated, this means that every valid string must end with a character with code 0. This character tells every function that is dealing with this string that it is in fact over.
In your program you create a string, signed char str[256]; and it is initially filled with random data; this means that you reserved space for 256 characters and they are all garbage, but the system does not know they are invalid. Try printing this string and see what happens.
In order to actually tell the system that your string is over after say, 8 characters, the 9th character hast to be the NUL character, or simply 0. In your code you can do it in two ways:
after the loop, assign str[i] = 0, or (even simpler)
initialize the string as signed char str[256]={0};, whiche creates the storage and fills it with nulls; after writing to the string you can be sure that the character after the last one you've written will be a NUL.
At the end of your do {} while () loop, you need to set the character after the last character in your string to 0. This is the array index of the last character you want (i) plus one. This lets printf know where your string ends. (Otherwise, how could it know?)
initialize the str variable to NUL.
void main (void)
{
signed char str[256];
int dec,rest;
int i = -1;
int j;
memset( str, '\0', sizeof(str) );
printf ("write a decimal number : ");
scanf ("%d",&dec);
do
{
rest = dec % 2;
dec/= 2;
for (j=i;j>-1;--j)
str[j+1]=str[j];
str[0]=rest+48;
++i;
} while (dec!=0);
printf ("the binary representation of this number is %s",str);
}

Subscripted value nor array nor pointer nor vector

This is my program to convert a binary to a decimal value.
#include <stdio.h>
#include <string.h>
#include <math.h>
void con(){
unsigned long long int dec = 0, bin;
int i;
printf ("\n Binary : ");
scanf("%lld",&bin);
for (i = strlen(bin) - 1; i <= 0; --i){ // Warning in here
dec = dec + (bin[i] * pow (2, i)); // Error in here
}
printf(" Decimal : %lld",dec);
con();
}
int main(){
con();
return 0;
}
When i compile the code, this error shows up ," Subscripted value is neither array nor pointer nor vector". And this warning also ," Passing argument 1 of strlen makes pointer from integer without a cast".
Why am i getting these and how can i fix them?
There are a number of problems with your code.
Most important is the variable bin. You probably want it to be a string like "1001001001" but you define it as unsigned long long int. Instead you should do:
char bin[100];
scanf("%s", bin); // Note: Not recommended! Use fgets instead.
Here scanf is not recommended as the user may overflow your buffer. Please use fgets instead.
As suggested by #MayurK: If you want to use scanf then at least do:
scanf("%99s", bin);
to prevent buffer from overflow.
Then this part:
dec = dec + (bin[i] * pow (2, i));
is wrong as bin[i] is not a number but a char.
You could do:
dec = 2 * dec + (bin[i] - '0'); // Note: No error checks which is bad
It will work as long as the user only inputs 0 and 1. In real code you should check that the user actually did so.
Finally you should not call con at the end of the function as it will give an endless loop. So delete that call:
printf(" Decimal : %lld",dec);
// DELETE THIS con();
}
dec = dec + (bin[i] * pow (2, i)); //bin is a numeric type
In this line you are trying to use a variable of unsigned long long type as array. Every numeric type is taken as a full value, you can't use array indexing to access separate digits. If you want to use it as a array use a char* or character array.
And you are also passing a unsigned long long to strlen(const char * str) function.

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