程式撞车的以下代码有什么错? 给予分割断层错误。 我使用 gcc 。
uint8_t result = 1
InsertRow("Name","Details of work",result);
void InsertRow(char *Name, char *Description,uint8_t Result)
{
char Buffer[500];
if(Result==1)
sprintf(Buffer,"<tr><td>%s </td> <td> %s </td> <td> %s </td></tr>",Name,Description,Result);
}
您重新使用 < code%s 格式化参数来表示类型 < code> beint8_t 的参数, 这应该是 < code_ u , 您应该将值投到 < code> untignign int 匹配。 这样可以避免您只关心确切类型并调整事项( 如评论者建议的那样) 。
此外,我们很难知道缓冲区足够大。当然,如果你有,你可以使用snprinf () 来避免。
给
sprintf(Buffer,"<tr><td>%s </td> <td> %s </td> <td> %s </td></tr>",Name,Description,Result);
您正在通过 Result (类型为 unit8_t ) 作为字符阵列的指针
这意味着集成值将作为一个指针被重新复制为指针, 该指针极有可能指向您无法访问的内存 -- 因此, segu. 错误 。 您需要用适当的格式标志替换第三个
note : 在此情况下, 不要直接使用 beint8_t 类型不能保证与 (很可能不是) 的大小相同。 如果您将 Result 的值投放到第一个 ( >(int)Result ), 您可以使用
结果不是一个字符*, 您可能想要%d 的字符*
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