Answer:
d)anotherFunc(myints);
Explanation:
When we call a function we just have to pass the names of the arguments to the function.In case of arrays we also just have to pass the name of the array.We don't have to worry about the square brackets.So the function call will be like this.
anotherFunc(myints);
Hence the answer is option d.
Solution :
#include
#include
#include
//Converts
to binary string.
* hexadecimal
Binary(char* hexdec)
{
long
= 0;
char *string =
(sizeof(char) * 9);
while (hexdec[i]) {
//Simply assign binary string for each hex char.
switch (hexdec[i]) {
strcat(string, "0000");
break;
strcat(string, "0001");
break;
strcat(string, "0010");
break;
strcat(string, "0011");
break;
strcat(string, "0100");
break;
strcat(string, "0101");
break;
strcat(string, "0110");
break;
strcat(string, "0111");
break;
strcat(string, "1000");
break;
strcat(string, "1001");
break;
case 'A':
case 'a':
strcat(string, "1010");
break;
case 'B':
case 'b':
strcat(string, "1011");
break;
case 'C':
case 'c':
strcat(string, "1100");
break;
case 'D':
case 'd':
strcat(string, "1101");
break;
case 'E':
case 'e':
strcat(string, "1110");
break;
case 'F':
case 'f':
strcat(string, "1111");
break;
default:
printf("\nInvalid hexadecimal digit %c",
hexdec[i]);
string="-1" ;
}
i++;
}
return string;
}
int main()
{ //Take 2 strings
char *str1 =hexadecimalToBinary("FA") ;
char *str2 =hexadecimalToBinary("12") ;
//Input 2 numbers p and n.
int p,n;
scanf("%d",&p);
scanf("%d",&n);
//keep j as length of str2
int j=strlen(str2),i;
//Now replace n digits after p of str1
for(i=0;i<n;i++){
str1[p+i]=str2[j-1-i];
}
//Now, i have used c library strtol
long ans = strtol(str1, NULL, 2);
//print result.
printf("%lx",ans);
return 0;
}
Answer:
Have a place for the who team can access all the information they need and use to put their work for project managers to check-up on. Having a when everything is organized in one place, the work will be much smoother.
Answer:
The governor found a way to free Sostre without assessing whether or not he was guilty or innocent of drug crime in buffalo.
Explanation: