警告:格式'%d'需要类型'int',但参数4的类型为'size_t'

Question

我按照一些类似的帖子来修改代码,但是仍然会生成警告.

$ g++ ncfile.c -o ncfile -g -lnetcdf

ncfile.c: In function ‘int main(int, char**)’:
ncfile.c:363: warning: format ‘%d’ expects type ‘int’, but argument 4 has type ‘size_t’
ncfile.c:363: warning: format ‘%d’ expects type ‘int’, but argument 4 has type ‘size_t’
ncfile.c:364: warning: format ‘%d’ expects type ‘int’, but argument 4 has type ‘size_t’
ncfile.c:364: warning: format ‘%d’ expects type ‘int’, but argument 4 has type ‘size_t’

在第363和364行的那个区块中:

  for(i = 0; i < ndimsp; i++) {
    char * temp_name;
    size_t temp_len;
    temp_name = (char *)malloc(sizeof(char) * 20);
    nc_inq_dim(cid, dimids[i], temp_name, &temp_len);
    dimlens[i] = temp_len;
    if(dimids[i] == unlimdimidp) printf("\t\t%d %s   \tlength: %d (UNLIMITED)\n", i, temp_name, temp_len);
    else printf("\t\t%d %s   \tlength: %d\n", i, temp_name, temp_len);
    total_records *= temp_len;
    free(temp_name);
  }

我该怎么摆脱这些警告.它对结果有害吗？

谢谢,

迈克尔

Answer 1

尝试使用z修饰符.对于size_t值,基本上是%zu.

这将是结果:

printf("\t\t%d %s   \tlength: %zu\n", i, temp_name, temp_len);

看看这个问题:

如何使用printf系列便携地打印size_t变量？