#include <stdio.h>
#include <math.h>
int convert(char *input)
{
//variable declaration and initialization
int res = 0;
int i, j=0;
char num[10];
char ch;
int error=0;
int count=0;
for (i = 0; input[i] != '\0'; i++)
{
ch = input[i];
if(isdigit(input[i]))
{
num[j] = ch;
j++;
}
else
{
error++;
}
}
i=j;
for(j=0; j<i; j++)
{
res = res + (num[j]-48) * pow(10, i-j-1);
}
//return statement
return res;
}
int main()
{
//character array declaration
char input[] = "1247";
int num;
//function calling
num = convert(input);
printf("%d", num);
return 0;
}
OUTPUT:
In C. // Extract the integer from the input string and convert to int int convert(char...
C code only! Complete this function so that the function str2int convert the char array s into its equivalent integer value. The char array has only digit characters of{0, 1, .., 9}and ’\0’.And the first element is not 0, which means, the integer string in the char array is of base 10. For example, if we pass chars[] = ”123” intostr2int, then it should return 123.Close the braces when you are done int str2int(char s[]){
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