小编abc*_*xyz的帖子

以下函数检查变量名称是否以字母开头,并且可能具有字母/数字的前面的字符.无论输入是什么,为什么返回值始终为1？

#include <regex.h>
#include <stdio.h>

int validate_var(char *str)
{
    regex_t reg;
    regcomp(&reg, "^[a-zA-Z]+[a-zA-Z0-9]*$", 0);
    int r = regexec(&reg, str, 0, NULL, 0);
    regfree(&reg);

    return r;
}

int main() {
    printf("%d\n", validate_var("abc")); // Reports 1, This makes sense
    printf("%d\n", validate_var("17"));  // Reports 1, This doesn't make sense
}

考虑一个指向结构的指针数组.以下代码取自您可能在此处找到的示例.我想对这两排铸造进行解释.我不熟悉这种"双重铸造".

int myptrstructcmp(const void *p1, const void *p2)
{
    struct mystruct *sp1 = *(struct mystruct * const *)p1;
    struct mystruct *sp2 = *(struct mystruct * const *)p2;

我认为它应该是:

int myptrstructcmp(const void *p1, const void *p2)
{
    struct mystruct *sp1 = (struct mystruct *)p1;
    struct mystruct *sp2 = (struct mystruct *)p2;

请考虑以下代码:

#include <stdio.h>

char** baz_alloc(int size)
{
    char ** b = malloc((size+1) * sizeof(char*));
    for (int i = 0; i < size; i++)
        b[i] = "baz";
    b[size] = NULL;

    return b;
}

void baz_print(char** baz)
{
    char** b = baz;
    while (*b != NULL)
    {
        printf("%s\n", *b);
        b++;
    }
}

void baz_free(char** baz)
{       
    int len = 0;
    char** b = baz;
    for (; *b != NULL; len++, b++);

    for (int i = 0; i < len; i++)
        free(baz[i]);
    free(baz); …