int main(void) { int a; char *s; int v0 = 9, v1 = 0, v2 = 0, v3 = 0; printf("Exercise 1:\n====
Run the following code and answer the questions.
#include <stdio.h>
int main(void) {
int a;
char *s;
int v0 = 9, v1 = 0, v2 = 0, v3 = 0;
printf("Exercise 1:\n====================\n");
switch(v0) {
case 0: printf("Good day\n"); break;
case 1: printf("Hey There!\n"); break;
case 2: printf("No Name\n"); break;
case 3: printf("Yes Please\n"); break;
case 4: printf("Go ");
case 5: printf("Vamos!\n");
default: printf(" "); break;
}
for(v1; v1<v2; v1=v1+v1) {
printf("Vamos ");
}
printf("\n");
if (v2 == 6) {
s = "Go";
}
if(v3 != 0) {
printf("%s VAMOS!\n",s);
}
return 0;
}
- What is the output of the
program ? Please explain why using the variables v0, v1, v2, and v3. - The code has a vulnerability that can prevent the program from outputting properly. Explain what the problem is. What are the values of v0, v1, v2, and v3 that may cause such problem?
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