C++ GnuCobol 使用 double 类型的参数调用 C 函数

我让C和C两个函数做完全相同的事情.唯一的区别是它们的返回类型和第一个参数类型:

int func1(int a, int *b) {
    if (a > 0) {
        *b = 0;
        return 1;
    }

    *b = -1;
    return 0;
}

double func2(double a, int *b) {
    if (a > (double)0.0) {
        *b = 0;
        return 1.0;
    }

    *b = -1;
    return (double)0.0;
}

将它们汇编到库中:

gcc -c -static -o testlib.o testlib.c

下面的COBOL代码调用这两个函数:

       IDENTIFICATION DIVISION.
       PROGRAM-ID. CallCFunctions.
       
       ENVIRONMENT DIVISION.

       DATA DIVISION.
       WORKING-STORAGE SECTION.
       01  VAR-AI USAGE BINARY-SHORT SIGNED.
       01  VAR-BI USAGE BINARY-SHORT SIGNED.
       01  VAR-CI USAGE BINARY-SHORT SIGNED.
       
       01  VAR-AD USAGE BINARY-DOUBLE.
       01  VAR-BD USAGE BINARY-DOUBLE.
       01  VAR-CD USAGE BINARY-DOUBLE.
       

       PROCEDURE DIVISION.
           
           MOVE 1 TO VAR-AI.
           MOVE -10 TO VAR-BI.
           MOVE ZERO TO VAR-CI.
           CALL "func1" USING BY VALUE VAR-AI BY REFERENCE VAR-BI
               RETURNING VAR-CI
           DISPLAY "Calling 'func1'".
           DISPLAY "A = ", VAR-AI, " B = ", VAR-BI, " C = ", VAR-CI

           MOVE 1 TO VAR-AD.
           MOVE -10 TO VAR-BD.
           MOVE ZERO TO VAR-CD.
           CALL "func2" USING BY VALUE VAR-AD BY REFERENCE VAR-BD
               RETURNING VAR-CD
           DISPLAY "Calling 'func2'".
           DISPLAY "A = ", VAR-AD, " B = ", VAR-BD, " C = ", VAR-CD


       STOP RUN.

编译COBOL代码:

cobc -x -free -o test test.cbl testlib.o

第一个函数可以工作，但第二个函数会生成错误:

try 引用未分配的内存(信号SIGSEGV) 全额输出

Calling 'func1'
A = +00001 B = +00000 C = +00001
Calling 'func2'

attempt to reference unallocated memory (signal SIGSEGV)

有人能解释一下问题出在哪里吗？