not sure what's going on in this code - c

Question

I have some C code, and I'm not quite sure what's going on.
#include <stdio.h>
#include <stdlib.h>
#define DIM1 7
#define DIM2 5
#define RES_SIZE 1000
typedef double stackElementT;
typedef struct {
stackElementT *contents;
int maxSize;
int top;
int min2;
} stackT;
void StackInit(stackT *stackP, int maxSize) {
stackElementT *newContents;
newContents = (stackElementT *)malloc(sizeof(stackElementT)*maxSize);
if (newContents == NULL) {
fprintf(stderr, "Not enough memory.\n");
exit(1);
}
stackP->contents = newContents;
stackP->maxSize = maxSize;
stackP->top = -1;
}
void StackDestroy(stackT *stackP) {
free(stackP->contents);
stackP->contents = NULL;
stackP->maxSize = 0;
stackP->top = -1;
}
int StackIsEmpty(stackT *stackP) { return stackP->top < 0; }
int StackIsFull(stackT *stackP) { return stackP->top >= stackP->maxSize-1; }
void StackPush(stackT *stackP, stackElementT element) {
if(StackIsFull(stackP)) {
fprintf(stderr, "Can't push element: stack is full.\n");
exit(1);
}
stackP->contents[++stackP->top] = element;
}
stackElementT StackPop(stackT *stackP) {
if(StackIsEmpty(stackP)) {
fprintf(stderr, "Can't pop element: stack is empty.\n");
exit(1);
}
return stackP->contents[stackP->top--];
}
int shell(char* s1, int arg) {
printf("> ");
scanf("%s %d%*c", &s1, &arg);
return arg;
}
int main() {
char cmds[DIM1][DIM2] = {{"push"}, {"pop"}, {"add"}, {"ifeq"}, {"jump"}, {"print"}, {"dup"}};
char* s1; int arg;
arg = shell(s1, arg);
printf("%s\n", &s1);
}
Input: push 4. It prints J+ instead of "push" but prints 4 normally.
It also gives these warnings on compile:
stack.c: In function ‘shell’:
stack.c:60: warning: format ‘%s’ expects type ‘char *’, but argument 2 has type ‘char **’
stack.c: In function ‘main’:
stack.c:71: warning: format ‘%s’ expects type ‘char *’, but argument 2 has type ‘char **’
stack.c:65: warning: unused variable ‘cmds’
stack.c:69: warning: ‘arg’ is used uninitialized in this function
Can someone please explain?

Answer 1

When you use the %s format specifier, it expect a value which is a pointer to the start of a string. In C, this type is char *.
Taking your main function, your variable s1 is of type char *. Therefore, s1 is a valid parameter to printf, so this line is valid:
printf("%s\n", s1);
Note the absence of an & in front of s1. In your code, you used the &, which takes the address of s1, the result of which will be of type char **. This is the wrong type, so don't use the &.
The thing is, printf can't actually tell what type its arguments are, since it is a variadic function. It simply uses whatever arguments are there, according to the types specified in the format string.
The same thing goes for scanf, but there is a pitfall: you must make sure that enough memory is allocated to account for the user input, else you will experience a buffer overflow with unpredictable results. Aside from this, printf and scanf are perfectly complementary.
Anyhoo, this takes care of the compiler warnings, aside from the unused cmds variable (it's unnecessary in the provided code). Also, there is the part of args - it really should be a variable declared inside of shell, and not passed as a parameter, since its value is not even used inside shell.
Don't know what's up with the rest of the code. It's superfluous considering your main function only calls on shell.