关于如何解决warning: assignment makes pointer from integer without a cast的警告
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关于如何解决warning: assignment makes pointer from integer without a cast的警告
原则:在源文件中没有发现函数的声明,可能是忘加头文件了。(未经声明的函数原型一律默认为返回int值)
这句话很重要,例如,CSDN上有这样一个问题:
http://topic.csdn.net/u/20080624/14/13953e05-da05-4598-a73e-8d81778599d1.html#r_achor
编写了段代码:
函数原形为:char *cfg_get(const char *name);
在main()中调用上述函数方式如下:
char *dev_IMEI = cfg_get((char *)"dm_imei");
但是此调用处编译出现warming如下:
warning: assignment makes pointer from integer without a cast
请问为啥会有这样的warming,如何修改才能不出现这样的warming?
这里应该要这么解决 char *dev_IMEI = (char *)cfg_get((char *)"dm_imei");
解释:这里由于cfg_get按照上面的原则该函数返回的是一个整形,而赋值的却是char *,所以需要强制转换下。同理,不同的类型进行不同的强制转换即可。
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