In this specific area of my project i am taking in info front he command line the doing different things. I'm getting this error even though the program executes correctly as is.
在我的项目的这个特定领域，我正在使用信息前端命令行执行不同的操作。即使程序按原样正确执行，我也会收到此错误。
lab2.c:86: error: conflicting types for ‘rrotate’
lab2.c:67: error: previous implicit declaration of ‘rrotate’ was here
here is the function and where it is called.
这是函数及其调用的位置。
CALL:
呼叫：
else if(strcmp(action, compare_r) == 0)
{
/* unsigned long rrotate(unsigned long x, int n)
this function will implement a right rotate function
that returns the value of the unsigned long x rotated to the
right by n bit positions*/
strcpy(e,argv[3]);
int n = strtol(e,0,10);
hex_num = (unsigned long)hex_one;
unsigned long number = rrotate(hex_num,n);
print_bits(number);
}
FUNCTION:
功能：
unsigned long rrotate(unsigned long x, int n){
unsigned long number = x >> n;
return number;
}
i run the program by compiling with gcc lab2.c -o lab2 then run like lab2 -r 0x5 3
我通过使用gcc lab2.c -o lab2编译运行程序，然后像lab2 -r 0x5 3一样运行
PROGRAM:
程序：
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
int main(int argc, char*argv[])
{
char action[5];
char compare_p[] = "-p";
char compare_i[] = "-i";
char compare_u[] = "-u";
char compare_c[] = "-c";
char compare_r[] = "-r";
char compare_s[] = "-s";
char compare_m[] = "-m";
char hex[10],e[5];
char *ptr;
unsigned long i,hex_num;
int hex_one,hex_two;
/*argc is the argument count-http://www.tutorialspoint.com/c_standard_library/c_function_atol.htm*//*It is the number of arguments passed into the program from the command line, including the name of the program.*/
if (argc < 3)
{
printf("NADA");
}
else
{
strcpy(action, argv[1]);
strcpy(hex,argv[2]);
hex_one = strtol(hex, &ptr, 16);
if(strcmp(action, compare_p) == 0)
{
print_bits(hex_one);
}
else if(strcmp(action, compare_u) == 0)
{
printf("do this instead");
}
else if(strcmp(action, compare_i) == 0)
{
/*create an intersection function*/
}
else if(strcmp(action, compare_c) == 0)
{/*complement funtion
0x7 -->> 1111 1111 1111 1111 1111 1111 1111 1000*/
hex_one = ~hex_one;
print_bits(hex_one);
}
else if(strcmp(action, compare_r) == 0)
{
/* unsigned long rrotate(unsigned long x, int n)
this function will implement a right rotate function
that returns the value of the unsigned long x rotated to the
right by n bit positions*/
strcpy(e,argv[3]);
int n = strtol(e,0,10);
hex_num = (unsigned long)hex_one;
unsigned long number = rrotate(hex_num,n);
print_bits(number);
}
else if(strcmp(action, compare_s) == 0)
{
}
else if(strcmp(action, compare_m) == 0)
{
}
}/*END ELSE*/
return 0;
}
unsigned long rrotate(unsigned long x, int n){
unsigned long number = x >> n;
return number;
}
print_bits(unsigned int i)
{
int x;
int count = 0;
for(x=(sizeof(int)*8)-1; x>=0; x--)
{
(i&(1<<x))?putchar('1'):putchar('0');
count+=1;
if(count == 4)
{
printf(" ");
count = 0;
}
}
printf("\n");
}
2 个解决方案
#1
1
Before the C99 standard, it was allowed to implicitly declare a function by calling without any previous declaration (function prototype). The compiler will then guess the functions arguments, by examining the types used in the call.
在C99标准之前，允许通过在没有任何先前声明（函数原型）的情况下调用来隐式声明函数。然后，编译器将通过检查调用中使用的类型来猜测函数参数。
However since the C99 standard this isn't really allowed, you need to declare a function before calling it. This can be very simply done by adding a function prototype before the call, so you should have something like
但是，由于C99标准并不是真正允许的，因此您需要在调用之前声明一个函数。这可以通过在调用之前添加函数原型来完成，所以你应该有类似的东西
// Function *declaration*
unsigned long rrotate(unsigned long x, int n);
int main(...)
{
...
}
// Function *definition*
unsigned long rrotate(unsigned long x, int n)
{
...
}
#2
1
Move the definition of rrotate before the main function or add a declaration before the main function. Your compiler actually declares rrotate when you call it.
在main函数之前移动rrotate的定义或在main函数之前添加声明。当你调用它时，你的编译器实际上声明了rrotate。
#1
1
Before the C99 standard, it was allowed to implicitly declare a function by calling without any previous declaration (function prototype). The compiler will then guess the functions arguments, by examining the types used in the call.
在C99标准之前，允许通过在没有任何先前声明（函数原型）的情况下调用来隐式声明函数。然后，编译器将通过检查调用中使用的类型来猜测函数参数。
However since the C99 standard this isn't really allowed, you need to declare a function before calling it. This can be very simply done by adding a function prototype before the call, so you should have something like
但是，由于C99标准并不是真正允许的，因此您需要在调用之前声明一个函数。这可以通过在调用之前添加函数原型来完成，所以你应该有类似的东西
// Function *declaration*
unsigned long rrotate(unsigned long x, int n);
int main(...)
{
...
}
// Function *definition*
unsigned long rrotate(unsigned long x, int n)
{
...
}
#2
1
Move the definition of rrotate before the main function or add a declaration before the main function. Your compiler actually declares rrotate when you call it.
在main函数之前移动rrotate的定义或在main函数之前添加声明。当你调用它时，你的编译器实际上声明了rrotate。