将二进制字符串转换为整数

Question

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

Answer 1

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];

Answer 2

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)"

Answer 3

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;
}