C 函数的隐式声明

Question

我使用的是 Linux 和 gcc 4.2.3。

对于下面的代码部分，隐式调用 lp_parm_talloc_string 函数，然后定义它：

char *lp_parm_string(const char *servicename, const char *type, const char *option)
{
        return lp_parm_talloc_string(lp_servicenumber(servicename), type, option, NULL);
}

/* Return parametric option from a given service. Type is a part of option before ':' */
/* Parametric option has following syntax: 'Type: option = value' */
/* the returned value is talloced in lp_talloc */
char *lp_parm_talloc_string(int snum, const char *type, const char *option, const char *def)
{
        param_opt_struct *data = get_parametrics(snum, type, option);

        if (data == NULL||data->value==NULL) {
                if (def) {
                        return lp_string(def);
                } else {
                        return NULL;
                }
        }

        return lp_string(data->value);
}

对于这部分，出现以下错误：

param/loadparm.c:2236: error: conflicting types for 'lp_parm_talloc_string'
param/loadparm.c:2229: error: previous implicit declaration of 'lp_parm_talloc_string' was here

如何告诉编译器允许这种情况？

Answer 1

您需要在使用函数之前声明它：

char *lp_parm_talloc_string(int snum, const char *type, const char *option, const char *def);

char *lp_parm_string(const char *servicename, const char *type, const char *option)
{
    return lp_parm_talloc_string(lp_servicenumber(servicename), type, option, NULL);
}

// ...and the rest of your code

或者只需更改两个函数在源代码中出现的顺序。