在C语言中,遇到“function declaration isn't a prototype”的警告,通常意味着函数声明没有使用函数原型。以下是对该问题的详细解答: 解释什么是函数原型(function prototype): 函数原型是指在函数声明时明确指出函数的返回类型、函数名和参数类型。函数原型有助于编译器在编译阶段检查函数调用中的类型匹配问题,从而提高...
error: function declaration isn't a prototype [-Werror=strict-prototypes] | int (*test_config)(); 解决方案 这是由于函数定义在没有参数的时候没有指定为void 将int (*test_config)();修改为 int (*test_config)(void);即可版权声明:本文为wgl307293845原创文章,遵循 CC 4.0 BY-SA 版权协议,转载请...
error:functiondeclarationisn’taprototype[-Werror=strict-prototypes]uint32_tsysctl_get_p() 原因:之前用的arm4.4.3编译链没报错,后面用的gcc-linaro6.5.0报错,看来是不同编译链的规则导致。函数无参数,不能空着,需要加void: uint32_t sysctl_get_p(void)...
Q: function declaration isn't a prototype A: http://hi.baidu.com/%D4%BC%D0%DE%D1%C7ing/blog/item/d5f86a0f598b47c27acbe138.html static void cut_mode(void) 如果写成static void cut_mode() 就会有警告function declaration isn't a prototype...
"warning: function declaration isn't a prototype" was caused by the function like that: return_type XXX() { ... } Please just modify the input paramter to void.return_type XXX(void) { ... } Or turn off the warning with -Wno-strict-prototypes (or simply by omitting -Wstrict-prototype...
function declaration isn't a prototype解决办法,即使函数括号内没有任何参数,也要加一个void类型,来避免这种warning:解决如下:
warning: function declaration isn’t a prototype(函数声明不是原型)的解决办法 linux驱动中定义一个无参的函数 int probe_num() { ... } 警告:函数声明不是一个原型 [-Wstrict-prototypes] 应对方法: 改成 int probe_num( void) { ... } 警告...
./ir.c:45: warning: function declaration isn't a prototype 虽然不是error,但总不顺眼.应该如何...
Function declaration isn't a prototype [-Wstrict-prototypes]espressif/arduino-esp32#3591 Closed TD-eradded a commit to TD-er/ESPEasy that referenced this issueDec 22, 2019 [PIO] Use -Wstrict-prototypes for C warnings c8da61c TD-ermentioned this issueDec 22, 2019 ...
/u01/oracle/11/rdbms/public/ociap.h:10794: warning: function declaration isn’t a prototype creating build/lib.linux-x86_64-2.6 gcc -pthread -shared -Wl,-O1 -Wl,-Bsymbolic-functions build/temp.linux-x86_64-2.6/cx_Oracle.o -L/u01/oracle/11/lib -L/u01/oracle/11 -lclntsh -o build/...