如下代码，为什么会出现这个错误呢？求解释！

Question

#include <stdio.h>
#include <math.h>
#include <ctype.h>

#define NUMBER '0'
#define MAXLEN 1000
#define BUFSIZE 100

int getch(void);
void ungetch(int);

int getline(char *);
void push(double );
double pop(void );
void ungets(char *);
double atof(char s[]);

char buf[BUFSIZE];
int bufp = 0;

#define MAXVAL 100

int sp = 0;
double val[MAXVAL];

main(int argc,char *argv[])
{
char s[MAXLEN];
double op2;

while(--argc > 0)
{
ungets(" ");
ungets(*++argv);

switch(getline(s))
{
case NUMBER : 
push(atof(s));
break;
case '+' :
push(pop() + pop());
break;
case '-' :
op2 = pop();
push(pop() - op2);
break;
case '*' :
push(pop() * pop());
break;
case '/' :
op2 = pop();
if(op2 != 0.0)
push(pop() / op2);
else 
printf("error : zero divisor \n");
break;
default :
printf("erro : unkown command %s\n",s);
argc = 1;
break;
}
}
printf("\t%.8g\n",pop());
return 0;
}
int getop(char s[])
{
int i,c;
while((s[0] = c =getch()) == ' ' || c == '\t')
;
s[1] = '\0';
if( !isdigit(c) && c != '.')
return c;
i = 0;
if(isdigit(c))
while(isdigit(s[++i] = c = getch()))
;
if(c == '.')
while(isdigit(s[++i] = c =getch()))
;
s[i] = '\0';
if(c != EOF)
ungetch(c);
return NUMBER;
}
int getch(void)
{
return (bufp > 0) ? buf[--bufp] : getchar();
}

void ungeth(int c)
{
if(bufp >= BUFSIZE)
printf("ungetch : too many characters\n");
else
buf[bufp++] = c;
}
void push(double f)
{
if(sp < MAXVAL)
val[sp++] = f;
else
printf("error: stack full , can't push %g\n");
}

double pop()
{
if(sp > 0)
return val[--sp];
else
{
printf("error: stack empty\n");
return 0.0;
}
}
double atof(char s[])
{
double val,power;
int i,sign;

for(i = 0; isspace(s[i]);i++)
;
sign = (s[i] == '-') ? -1 : 1;

if(s[i] == '+' ||s[i] == '-')
i++;
for(val = 0.0;isdigit(s[i]);i++)
val = 10.0 * val + s[i] - '0';
if(s[i] == '.')
{
i++;
for(power = 1.0 ;isdigit(s[i]);i++)
{
val = 10.0 * val + s[i] - '0';
power *= 10.0 ;
}
}
return sign * val / power;
}

Answer 1

有重复声明的函数，如double atof(char s[])，把这个改个名，这个告警不会有了

Answer 2

没事的，你的程序没有错误