找不到 "Too many arguments for format"的问题

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已关闭。这个问题是 not reproducible or was caused by typos 。目前不接受答案。

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这个问题是由拼写错误或无法再重现的问题引起的。虽然类似的问题可能是 on-topic在这里，这个问题的解决方式不太可能帮助 future 的读者。

已关闭 4 年前。

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我的 sprintf 语句遇到问题。我看到 6 个参数格式和 6 个提供的参数，但出现以下错误:

jsonServer.c:450:4: error: unknown conversion type character ‘}’ in format [-Werror=format=]
sprintf(message, "{\"num_clients\": %d,\"num_requests\": %d,\"errors\": %d,\"uptime\": %u,\"cpu_time\": %lu,\"memory_used\": %l}", (numConnections+1), numRequests, numErrors, uptime, cpuTime, memUsed);
jsonServer.c:450:4: error: too many arguments for format [-Werror=format-extra-args]

char *buildStatus()
{
   struct rusage *usage = malloc(sizeof(struct rusage));
   int usageRet = getrusage(RUSAGE_SELF, usage);
   if (usageRet == -1)
   {
      perror("RUSAGE fail");
      exit(EXIT_FAILURE);
   }
   long unsigned cpuTime = (usage->ru_utime).tv_sec + (usage->ru_stime).tv_sec;
   long memUsed = get_memory_usage_linux();
   unsigned int uptime = 0;

   char *message = malloc(1000);
   sprintf(message, "{\"num_clients\": %d,\"num_requests\": %d,\"errors\": %d,\"uptime\": %u,\"cpu_time\": %lu,\"memory_used\": %l}", (numConnections+1), numRequests, numErrors, uptime, cpuTime, memUsed);
   free(usage);
   return message;
}

我认为存在一些偷偷摸摸的转义字符问题，但是在到处粘贴反斜杠之后，我似乎无法修复它。

Answer 1

您所要做的就是更正%l(没有这样的说明符)，您可能应该使用%ld。

char *buildStatus()
{
   struct rusage *usage = malloc(sizeof(struct rusage));
   int usageRet = getrusage(RUSAGE_SELF, usage);
   if (usageRet == -1)
   {
      perror("RUSAGE fail");
      exit(EXIT_FAILURE);
   }
   long unsigned cpuTime = (usage->ru_utime).tv_sec + (usage->ru_stime).tv_sec;
   long memUsed = get_memory_usage_linux();
   unsigned int uptime = 0;

   char *message = malloc(1000);
   sprintf(message, "{\"num_clients\": %d,\"num_requests\": %d,\"errors\": %d,\"uptime\": %u,\"cpu_time\": %lu,\"memory_used\": %ld}", (numConnections+1), numRequests, numErrors, uptime, cpuTime, memUsed);
   free(usage);
   return message;
}

希望有帮助。